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Business Mathematics MCQS
Chapter 1: Ratio and Proportion, Indices, Logarithms
Question 1.
Two numbers are in the ratio of 2 : 3 and the difference of their squares is 320. The number are :
(a) 12, 18
(b) 16, 24
(c) 14, 21
(d) None
Answer:
Tricks : Go by choices (a); (b) & (c) all are in ratio 2:3 But For option
(a) 182 – 122 ≠ 320
(b) 242 – 162 = (24+ 16) (24 – 16)
= 40 × 8 = 320
Question 2.
If p : q is the sub-duplicate ratio of p – x2: q – x2, then x2 is :
(a) pp+q
(b) qp+q
(c) qpp−q
(d) None
Answer:
Detail Method:
p−x2√q−x2√=pq
Squaring on both side; we get
p−x2q−x2=p2q2
or pq2 – q2x2 = p2q – p2x2
or p2x2 – q2x2 = p2q – pq2
or x2(p2 – q2) = pq(p – q)
or x2 (p + q)(p-q) = pq (p – q)
(d) is correct
Tricks : Go by choices.
Question 3.
An alloy is to contain copper and zinc in the ratio 9:4. The zinc required to melt with 24 kg of copper is :
(a) 1023kg
(b) 1013kg
(c) 923kg
(d) 9 kg
Answer:
Let Zinc = x kg
(a) ∴ 94=24x ∴ x = 4×249=323
= 1023kg
∴ (a) is correct
Question 4.
Two numbers are in the ratio 7: 8. If 3 is added to each of them, their ratio becomes 8 : 9. The numbers are:
(a) 14,16
(b) 24,27
(c) 21,24
(d) 16,18
Answer:
Tricks : Go by choices
(b) and (d) are not in the ratio 7 : 8 So (b) & (d) are not answer For (a) it is added then
14+310+3=1719≠89
(a) is not answer
(c) is answer Detail Method:
Let x is common in the ratio
∴ Numbers are 7x & 8x
Now 7x+38x+3=89
or 64x + 24 = 63x + 27
or 64x – 63x = 27-24
or x = 3
1st number = ix = 7×3 = 21
2nd number = 8x = 8×3 = 24
(c) is correct
Question 5.
A box contains ₹ 56 in the form of coins of one rupee, 50 paise and 25 paise. The number of 50 paise coin is double the number of 25 paise coins and four times the numbers of one rupee coins. The numbers of 50 paise coins in the box is :
(a) 64
(b) 32
(c) 16
(d) 14
Answer:
Tricks : Go by choices
No. of 25 paise coins = 12 No. of 50 Paise
No. of Rupee coins = 14 No. of 50 Paise Coins
For (a) No. of coins of ₹ 1;50 Paise & 25 Paise
644; 64; 642
Total Value = 16 × 1 + 64 × 0.50 + 32 × 0.25
= ₹ 16 + 32 + 8 = ₹ 56
Which is equal to given value (a) is Correct
Detail Method
Let No. of 50 Paise coins = x
∴ No. of ₹ 1 coins = x4
and No. of 25 Paise coins = x2
∴ Total Value =
x4 × 1 + x × 0.50+ x2 × 0.25 = 56
or 0.25x + 0.50x + 0.125x
or 0.875x = 56
or x = 560.875 = 64
∴ (a) is correct
Question 6.
Eight people are planning to share equally the cost of a rental car. If one person withdraws from the arrangement and the others share equally entire cost of the car, then the share of each of the remaining persons increased by :
(a) 1/9
(b) 1/8
(c) 1/7
(d) 7/8
Answer:
Tricks : Per Person share increase = 17 of total share
Question 7.
A bag contains ₹ 187 in the form of 1 rupee, 50 paise and 10 paise coins in the ratio 3:4:5. Find the number of each type of coins:
(a) 102, 136, 170
(b) 136, 102, 170
(c) 170, 102, 136
(d) None
Answer:
Tricks I: Go by choices
For (a) Coins are in the ratio 3:4:5
∴ 1023=1364=1705 = 34
It satisfies 1st condition
Now 102 × 1 + 136 × 0.50 + 170 × 0.10
= 102 + 68 + 17 = ₹ 187
∴ (a) is correct
Tricks II
Common factor = 1873×1+4×0.50+5×0.10
= 1875.50 = 34
∴ No. of 1 Rupee coins =3 × 34 = 102
No. of 50 Paise coins =4 × 34 = 136
No. of 10 paise coins = 5 × 34 = 170 (a) is correct
Detail Method
Let x is common in the ratio
No. of 1 Rupee ; 50 Paise and 10 Paise Coins are 3x ; 4x and 5x
∴ 3x × 1 + 4x × 0.50 + 5x × 0.10 = 187
or 5.50x = 187
∴ x = 1875.50 = 34
∴ No. of 1 Rupee coins = 3x = 3 × 34 = 102
No. of 50 Paise coins = 4x = 4 × 34 = 136
No. of 10 Paise coins = 5x = 5 × 34 = 170
∴ (a) is correct
Question 8.
Ratio of earnings of A and B is 4 : 7. If the earnings of A increase by 50% and those of B decrease by 25%, the new ratio of their earning becomes 8 : 7. What is A’s earning ?
(a) ₹ 21,000
(b) ₹ 26,000
(c) ₹ 28,000
(d) Data inadequate
Answer:
Detailed Method Let x is common in the ratio
∴ A’s and B’s present earnings are 4x and 7x respectively
From question 4x+4x×0.507x−7x×0.25=87
or 6x5.25x=87
x cannot be found.
Data is inadequate
∴ (d) is Correct
Question 9.
P, Q and R are three cities. The ratio of average temperature between P and Q is 11 : 12 and that of between P and R is 9: 8. The ratio between the average temperature of Q and R is :
(a) 22: 27
(b) 27: 22
(c) 32: 33
(d) None
Answer:
∴ P: Q= 11 : 12 ∴ Q :P= 12 : 11
QP×PR=1211×98
QR=2722 Q :P= 12 : 11
(b) is Correct
Question 10.
₹ 407 are to be divided among A, B and C so that their shares are in the ratio 14:15:16
The respective shares of A, B, C are:
(a) ₹ 165, ₹ 132, ₹ 110
(b) ₹ 165, ₹ 110, ₹ 132
(c) ₹ 132, ₹ 110, ₹ 165
(d) ₹ 110, ₹ 132, ₹ 165.
Answer:
A:B:C = 14:15:16 × LCM of denominators = 60
= 15 : 12 : 10
∴ A’s share = 40715+12+10 × 15 = Rs. 165
B’s share = 40737 × 12 = Rs. 132
C’s share = 40737× 10 = Rs. 110
(a) is Correct
Tricks : Go by Choices.
Question 11.
The incomes of A and B are in the ratio 3 :2 and their expenditures in the ratio 5 : 3. If each saves ₹ 1,500, then B’s income is :
(a) ₹ 6,000
(b) ₹ 4,500
(c) ₹ 3,000
(d) ₹ 7,500
Answer:
Detail Method Let x is common in the ratio.
A’s income = 3x
B’s income = 2x
∴ 3x−15002x−1500=53
or 10x – 7500 = 9x – 4500
or 10x – 9x = 7500 – 4500
or x = 3000
B’s income = 2x =2 × 3000
= ₹ 6000.
(a) is Correct Tricks : Go by choices For (a)
AB Expenditure = 60002×3−15006000−1500=75004500=53
(a) is Correct
Question 12.
In 40 litres mixture of glycerine and water, the ratio of glycerine and water is 3:1. The quantity of water added in the mixture in order to make this ratio 2:1 is :
(a) 15 litres
(b) 10 litres
(c) 8 litres
(d) 5 litres
Answer:
Glycerine = 403+1 x 3 = 30 litres.
Water = 404 x 1 = 10 litres 4
Let x litres of water is added to the mixture
Then 3010+x=21
or, 2x + 20 = 30
or x = 5
∴ (d) is Correct
Tricks : Go by Choices
Question 13.
The third proportional between (a2 – b2) and (a + b)2 is :
(a) a+ba−b
(b) a−ba+b
(c) (a−b)2a+b
(d) (a+b)3a−b
Answer:
3rd Proportion = ( Mean prop. )2 lst Proportional
= {(a+b)2}2a2−b2=(a+b)43(a+b)(a+b)=(a+b)3a−b
(d) is Correct
Question 14.
In what ratio should tea worth ₹ 10 per kg. be mixed with tea worth ₹ 14 per kg., so that the average price of the mixture may be ₹ 11 per kg.?
(a) 2 : 1
(b) 3 : 1
(c) 3 : 2
(d) 4 : 3
Answer:
∴ (b) is Correct
Question 15.
The ages of two persons are in the ratio 5:7. Eighteen years ago their ages were in the ratio of 8:13, their present ages (in years) are :
(a) 50; 70
(b) 70,50
(c) 40,56
(d) None
Answer:
Tricks : Go by choices (a) & (c) are in the ratio 5 : 7 not (b)
For (a) 18 year ago
So, (a) is Correct
Question 16.
If A, B and C started a business by investing ₹ 1,26,000, ₹ 84,000 and ₹ 2,10,000. If at the end of the year profit is ₹ 2,42,000 then the share of each is : [1 Mark, Dec. 2008]
(a) ₹ 72,600; ₹ 48,400 ; ₹ 1,21,000
(b) ₹ 48,400 ; ₹ 1,21,000 ; ₹ 72,600
(c) ₹ 72,000 ; ₹ 49,000 ; ₹ 1,21,000
(d) ₹ 48,000 ; ₹ 1,21,400 ; ₹ 72,600
Answer:
Investment ratio is
A : B : C = 126,000 : 84,000 : 2,10,000 ÷ 14,000
= 9: 6 : 15 ÷ 3
= 3: 2: 5
A’s share = ₹242,0003+2+5 × 3 =₹ 72, 600
B’s share = 242,00020 × 2 = ₹48, 400
C’s share = 24200010 × 5 = ₹ 1,21,000
So, (a) is Correct
Question 17.
If pq=−23 then the value of 2p+q2p−q is:
(a) 1
(b) −17
(c) 17
(d) 7
Answer:
∵ pq=−23
Tricks
2p+q20−q=2(−2)+32(−2)−3=−4+3−4−3=−1−7=17
(c) is correct
Question 18.
Fourth proportional to x, 2x, (x +1) is :
(a) x + 2
(b) (x + 2)
(c) (2x + 2)
(d) (2x – 2)
Answer:
Let Fourth Proportional is K.
∴ x2x=x+1K
or k.x = 2x (x + 1)
or k = 2 (x + 1) = 2x +2
(c) is correct
Question 19.
What must be added to each term of the ratio 49 : 68 so that it becomes 3 : 4?
(a) 3
(b) 5
(c) 8
(d) 9
Answer:
Detail Method:
Let x is added to each term
Then 49+x68+x=34
or 196 +4x = 204 + 3x
or 4x – 3x = 204 – 196
or x = 8
(c) is Correct
Tricks : Go by Choices
1st Find 34 = 0.75 (By Calculator)
Question 20.
The students of two classes are in the ratios 5 : 7, if 10 students left from each class, the remaining students are in the ratio of 4 : 6, then the number of students in each class was :
(a) 30, 4
(b) 25, 24
(c) 40, 60
(d) 50, 70
Answer:
Tricks : Go by choices:
(a); (b) and (c) are not in the ratio 5 : 7
∴ (d) is Correct.
Question 21.
If A: B= 2: 5, then (10A + 3B) : (5A + 2B) is equal to: [1 Mark, Dec. 2010]
(a) 7: 4
(b) 7: 3
(c) 6: 5
(d) 7: 9
Answer:
It A : B = 2 : 5 Then
10A+3B5A+2B=10×2+3×55×2+2×5=3520=74
= 7: 4
(a) is Correct
Question 22.
In a film shooting, A and B received money in a certain ratio and B and C also received the money in the same ratio. If A gets ₹ 1,60,000 and C gets ₹ 2,50,000. Find the amount received by B ?
(a) ₹ 2,00,000
(b) ₹ 2,50,000
(c) ₹ 1,00,000
(d) ₹ 1,50,000
Answer:
Detail Method
A : B = B : C
So, B2 = AC ;
so, B = AC−−−√=1,60,000×2,50,000−−−−−−−−−−−−−−−−√
= 400 × 500 = 2,00,000
Question 23.
The ratio compounded of 4:5 and sub-duplicate of “ a” : 9 is 8:15. Then value of “a” is:
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 45×a9−−√=815
or 45×a√3=815
∴ √a = 2 ⇒ a = 4
(c) is Correct
Question 24.
If X varies inversely as square of Y and given that Y=2 for X = 1, then the value of X for Y =6 will be:
(a) 3
(b) 9
(c) 1/3
(d) 6
Answer:
(d) is Correct
x × 1y2 ⇒ x = K.1y2 x = ky2, where k = proportional constant
where k = proportional constant
When x = 1 Then y = 2
1 = k22 ⇒ k = 4 ∴ x = 4y2
When y = 6, Then x = 462=19
x = 19
Question 25.
Which of the numbers are not in proportion ?
(a) 6, 8, 5, 7
(b) 7, 14, 6, 12
(c) 18, 27,12, 18
(d) 8, 6, 12, 9
Answer:
(a) Go by choices
For (a) 68=34≠57
(a) is not in proportion
Question 26.
Find two numbers such that mean proportional between them is 18 and third proportional between them is 144:
(a) 9 ; 36
(b) 8 ; 32
(c) 7 ; 28
(d) 6 ; 14
Answer:
(a) is correct Tricks : Go by choices
For (a) Mean Proportional of 9 and 36
= 9×36−−−−−√ = 18
It satisfies 1st condition.
If 144 is its 3rd condition.
362 = 9 × 144
It also satisfies the 2nd Condition.
Question 27.
Triplicate ratio of 4 : 5 is:
(a) 125 : 64
(b) 16 : 25
(c) 64 : 125
(d) 120 : 46
Answer:
(c) Triplicate ratio of 4:5 = 43 :53 = 64: 125
Question 28.
The mean proportion between 24 and 54 is _______.
(a) 33
(b) 34
(c) 35
(d) 36
Answer:
(d) Mean – Proportion = 24×54−−−−−−√ = 36
Question 29.
The ratio of numbers is 1:2:3 and sum of their squares is 504 then the numbers are:
(a) 6, 12, 18
(b) 3, 6, 9
(c) 4, 8, 12
(d) 5, 10, 15
Answer:
(a) is correct
Tricks : Go by choices
Tricks : See Quicker BMLRS
Question 30.
If P is 25% less than Q and R is 20% higher than Q the Ratio of R and P:
(a) 5:8
(b) 8:5
(c) 5:3
(d) 3:5
Answer:
(b) is correct
Let Q = 100, So, P = 100 – 025 = 75
&R = 100 + 20= 120
RP=12075=85
Question 31.
A person has assets worth ₹ 1,48,200. He wish to divide it amongst his wife, son and daughter in the ratio 3:2:1 respectively. From this assets the share of his son will be:
(a) ₹ 74,100
(b) ₹ 37,050
(c) ₹ 49,400
(d) ₹ 24, 700
Answer:
(c) is correct
Share of son = 23+2+1 × 1,48,200
= ₹ 49,400
Question 32.
If x : y = 2 : 3 then (5x+2y): (3x -y) =
(a) 19 : 3
(b) 16: 3
(c) 7 : 2
(d) 7: 3
Answer:
(b) is correct
5x+2y3x−y=5×2+2×33×2−3=163
Question 33.
The first, second and third month salaries of aperson are in the ratio 2:4:5. The difference between the product of the salaries of first 2 months & last 2 months is ₹ 4,80,00,000. Find the salary of the second month
(a) ₹ 4,000
(b) ₹ 6,000
(c) ₹ 12,000
(d) ₹ 8,000
Answer:
(d) is correct
Let x is common in the ratio.
1st, 2nd and 3rd month salaries of a person = 2x ; 4x ; 5x
From Qts.
4x × 5x – 2x × 4x = 4,80,00,000.
or, 12x2 = 4,80,00,000.
or, x2 = 4000000
x = 2000.
2nd month salary = 4x = 4×2000
= ₹ 8000
Question 34.
(2p2 – q2) = 7pq, where p, q are positive then p : q.
(a) 5:6
(b) 5:7
(c) 3:5
(d) 3:7
Answer:
(a) is correct 15(2p2 – q2) = 7pq
Tricks : Go by choices
For (a) put p = 5; q = 6 we get
15[2 × 52 – 62] = 3 × 5 × 6
or 15 × 14 = 210
or 210 = 210
Question 35.
If one type of rice of cost ₹ 13.84 is mixed with another type of rice of cost ₹ 15.54, the mixture is sold at ₹ 17.60 with a profit of 14.6% on selling price then in which proportion the two types of rice mixed ?
(a) 3:7
(b) 5:7
(c) 7:9
(d) 9:1
Answer:
Cost of mixture per kg = 17.60 – 14.6% = 15.0304 = 15.03 (approx.)
By rules of Alligation
51: 119 = 3: 7
Go by choices
(a) is correct (approx)
Question 36.
Find the ratio of third proportional of 12 ; 30 and mean proportional of 9; 25 :
(a) 7: 2
(b) 5 : 1
(c) 9 : 4
(d) None of these
Answer:
3rd proportional = 30212 = 75
Mean Proportional = 9×25−−−−−√ =15
Ratio = 7515 = 5:1
(b) is correct
Question 37.
What must be added to each of the numbers 10, 18, 22, 38 to make them proportional:
(a) 5
(b) 2
(c) 3
(d) 9
Answer:
(b) is correct
let x be added.
∴ 10+x18+x=22+x38+x
Tricks: Go by choices.
∴ x = 2 satisfies it.
Question 38.
x, y, z together starts a business, if x invests 3 times as much as y invests and y invests two third of what z invests, then the ratio of capitals of x, y, z is:
(a) 3 : 9 : 2
(b) 6: 3: 2
(c) 3 : 6 : 2
(d) 6: 2: 3
Answer:(d)
Tricks: Go by choices
6 = 3 × 2 and 2 = 3 × 23
Question 39.
A bag contains 23 number of coins in the form of 1 rupee, 2 rupee and 5 rupee coins. The total sum of the coins is ?₹ 43. The ratio between 1 rupee and 2 rupees coins is 3 : 2. Then the number of 1 rupee coins is:
(a) 12
(b) 8
(c) 10
(d) 16
Answer:
(a)
Tricks : Go by choices
Let option (a) is correct.
Let x is common in the ratio.
So, ₹ 1 coins = 3x = 12 ; So, x = 4
No. of ₹ 2 coins = 2 × 4 = 8
Hence no. of coins of ₹ 5 coins = 23 – 12 – 8= 3
Total money = 12 × 1 + 8 × 2 + 3 × 5 = ₹ 43
Satisfied. So (a) is correct.
Question 40.
If a: b = 2: 3, b : c = 4: 5, c: d = 6: 7 then a : d is:
(a) 24 : 35
(b) 8 : 15
(c) 16 : 35
(d) 7 : 15
Answer:
Option (c) is correct.
Multiply all ratios.
= 23×45×67=1635
Question 41.
The ratio of the number of five rupee coins to number of ten rupee coins is 8: 15. If the total value of five rupee coins is 360, then the no. of ten rupee coins is _______.
Answer:
Option (d) is correct.
Total No. of ₹ 5 coins = 360/5 = 72
Let x is common in the ratio.
So, ₹ 5 coins = 8x = 72 ; So, x = 9
No. of ₹ 10 coins = 15 × 9 = 135
Question 42.
If 12,13,15,1x are in proportion then x = .
(a) 152
(b) 315
(c) 215
(d) 115
Answer:
Option (a) is correct.
Product of middle two terms = Product of extremes
So, 12x=115; x = 15/2
Question 43.
If (a + b): (b + c): (c + a) = 7 : 8 : 9 and a + b + c = 18 then a : b : c = .
(a) 5 : 4 : 3
(b) 3 : 4 : 5
(c) 4 : 3 : 5
(d) 4: 5 : 3
Answer:
(c) 4 : 3 : 5 is correct
Tricks: Go by choices.
(c) Let a : b : c = 4 : 3 : 5
It is in ratio. So, it should must satisfy given ratio (a + b): (b + c): (c + a) = 7 : 8 : 9
i.e. (4 + 3): (3 + 5): (5 + 4) = 7 : 8 : 9 (True) Avoid 2nd condition.
In detail it will take too much time.
Question 44.
If p : q is the sub-duplicate ratio of p – x2:q – x2, then x2 is :
(a) pp+q
(b) qp+q
(c) qpp−q
(d) None
Answer:
Question 45.
The mean proportional between 24 and 54 is :
(a) 33
(b) 34
(c) 35
(d) 36
Solution:
Formula
Mean Proportion of a & b = ab−−√
(d) = 24×54−−−−−−√ = 36
Question 46.
3x−25x+6 is the duplicate ratio of 23 then find the value of x :
(a) 6
(b) 2
(c) 5
(d) 9
Answer:
(a)
Given 3x−25x+6=(23)2=49
Tricks : Go by choices
for option (a) putting x = 6 in LHS; we get
3×2−25×2+6=49
∴ (a) is correct.
Question 47.
If x : y : z = 7 : 4 : 11 then is:
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a)
x+y+zz=7+4+1111 = 2
Question 48.
If the ratio of two numbers is 7 : 11. If 7 is added to each number then the new ratio will be 2 : 3 then the numbers are.
(a) 49,77
(b) 42,45
(c) 43,42
(d) 39,40
Answer:
Tricks:- GBC (Go by Choices)
(a)
49 ÷ 7 = 7
77 ÷ 11 = 7 both must be equal.
Here it is correct.
Now:
49+777+7=5684=23
Divide 56 by numerator (2) and 84 by Denominator (3) we get same value “28”
Note:- No need to solve ; only check.
By Calculator.
Indices
Question 1.
Value of (a1/8 + a-1/8)(a1/8 – a-1/8)
(a1/4 + a-1/4)(a1/2 + a-1/2) is:
(a) a + 1a
(b) a – 1a
(c) a2 + 1a2
(d) a2 – 1a2
Answer:
[a1/8 + a-1/8][a1/8 – a-1/8][a1/4 + a-1/4][a1/2 + a-1/2]
[Use Formula (a + b)(a – b) = a2 – b2]
= [(a1/8)2 – (a1/8)2][a1/4 + a-1/4][a1/2 + a-1/2]
= (a1/4 – a-1/4)(a1/4 + a-1/4)(a1/2 + a-1/2)
= [(a1/4)2 – (a-1/4)2][a1/2 + a-1/2]
= (a1/2 – a-1/2)(a1/2 + a-1/2)
= (a1/2)2 – (a-1/2)2 = a – a-1 = a – 1a
(b) is correct
Question 2.
Simplification of xm+3nx4m−9nx6m−6n is :
(a) xm
(b) x-m
(c) xn
(d) x-n
Answer:
xm+3nx4m−9nx6m−6n = xm+3n+4m-9n-6m+6n = x-m = x-m
(b) is correct
Question 3.
On simplification 11+za−b+za−c+11+zb−c+zb−a+11+zc−a+zc−b reduces to : [1 Mark, Aug. 2007]
(a) 1z2(a+b+c)
(b) 1z(a+b+c)
(c) 1
(d) 0
Answer:
Tricks
11+Za−b+Za−c+11+Zb−c+Zb−a+11+Zc−a+Zc−a
= 1 [it is in cyclic order]
Question 4.
If 4x = 5y = 20z then z is equal to:
(a) xy
(b) x+yxy
(c) 1xy
(d) xyx+y
Answer:
(d)
Let 4y = 5y = 2oz =k
or 4 = k1/x; 5 = k1/y; 20 = k1/z
∴ 20 = 4 × 5
(d) is correct
Question 5.
(3√9)5/2(933√)7/2 × 9 is equal to: [1 Mark, Nov. 2007]
(a) 1
(b) √3
(c) 3√3
(d) 393√
Answer:
= (3-4)1/2.32 = 3-2.32 = 3-2+2 = 30 = 1
(a) is correct.
Question 6.
If 2x – 2x – 1 = 4, then the value of xx is : [1 Mark, Feb. 2008, June 2010]
(a) 2
(b) 1
(c) 64
(d) 27
Answer:
2x – 2x-1 = 4
or 2x-1 (2 – 1) = 4
or 2x-1 x1 = 22
or 2x-1 = 22
x – 1 = 2
x = 3
…xx = 33 =27
(d) is Correct
Tricks : Go by choices
For (d) 27 = 33 = xx
x = 3
Put x = 3 in 2x – 2x-1 = 4
It satisfies it (d) is correct
Question 7.
If x = ya, y = zb and z = xc then abc is : [1 Mark, June 2008]
(a) 2
(b) 1
(c) 3
(d) 4
Answer:
It x = ya; y = Zb & Z = xc
Then abc = 1
Tricks : It is in cyclic order.
(b) is correct
Tricks : See Quicker BMLRS example
Question 8.
If x = 31/3 + 3-1/3 then find value of 3x3 – 9x. [1 Mark, June 2009]
(a) 3
(b) 9
(c) 12
(d) 10
Answer:
Detail Method
It x = 31/3 + 3-1/3 ……….. (I)
Cubing on both sides; we get x3 =(31/3)3 + (3-1/3)3 +3.31/3.3-1/3(31/3 + 3-1/3)
= 3 + 3-1 + 3 × 1 × x
or x3 = 3 + 13 + 3x
or x3 – 3x = 9+13
or 3x2 – 9x = 10
∴ (d) is correct
Tricks See Quicker BMLRS for CA-Found.
Question 9.
Find the value of: [1 – {1 – (1 – x2)-1}-1]-1/2 is
(a) 1/x
(b) x
(c) 1
(d) none of these
Answer:
∴ (b) is correct
Question 10.
2n+2n−12n+1−2n
(a) 1/2
(b) 3/2
(c) 2/3
(d) 1/3
Answer:
∴ (b) is correct
Tricks: Put n = 1
Question 11.
If 2x × 3y × 5z = 360. Then what is the value of x, y, z. ? [1 Mark, Dec. 2009]
(a) 3, 2, 1
(b) 1,2,3
(c) 2, 3, 1
(d) 1, 3, 2
Answer:
If 2x × 3y × 5z = 360
2x × 3y × 5z = 23 × 32 × 51
Comparing it; we get
x = 3; y = 2; z = 1;
∴ (a) is correct
Tricks : Go by choices.
For (a) x = 3; y = 2; Z = 1
LHS= 23 × 32 × 51 =360 = RHS.
(a) is correct
Question 12.
The recurring decimal 2.7777 ………. can be expressed as: [1 Mark, Dec. 2010]
(a) 24/9
(b) 22/9
(c) 26/9
(d) 25/9
Answer:
Tricks : Go by choices.
By calculator
(a) 249 = 2.666 …………………. ≠ 2.777
(b) 229 = 2.444 ……….. ≠ 2.777
(c) 269 = 2.888 …………… ≠ 2.77
(d) 259 = 2.777
(d) is correct
Question 13.
The value of (3n+1+3n)(3n+3−3n+1) is equal to:
(a) 1/5
(b) 1/6
(c) 1/4
(d) 1/9
Answer:
(b) Tricks :
Put n = 0
Question 14.
Find the value of X, if x.(x)1/3 = (x1/3)z. [1 Mark, Dec. 2012]
(a) 3
(b) 4
(c) 2
(d) 6
Answer:
(b) is correct
x.x13 = xx3
or x1+13 = xx/3
∴ 1 + 13=x3
∴ 43=x3
∴ x = 4
(b) is correct.
Question 15.
If a−−√3+b√3+c√3 = 0; then find the value of [a+b+c3]3 =
(a) 9abc
(b) 19abc
(c) abc
(d) 1abc
Answer:
(c) is correct
Tricks:
Let a = -1; b= -1 and c =8, because a−−√3+b√3+c√3=−1−−−√3+−1−−−√3+8–√3
= -1 – 1 + 2 = 0(R.H.S) ∴ [a+b+c3]3=[−1−1+83]3 = (2)3 = 8
= (-1).(-1).(8) = abc
∴ (c) is correct
Question 16.
The value of
(yayb)a2+ab+b2(ybyc)b2+bc+c2(ycya)c2+ca+a2 = . [1 Mark, June 2014]
(a) y
(b) -1
(c) 1
(d) None
Answer:
(c) is correct
Tricks: Cyclic order
Question 17.
If px = q, qy = r, rz = p6, then the value of xyz is ……….[1 Mark, June 2015]
(a) 0
(b) 1
(c) 3
(d) 6
Answer:
qy = r ⇒ (px)y = r ⇒ r = pxy
Now rz = p6 ⇒ (pxy)z = p6 ⇒ pxyz = p6
xyz = 6
Question 18.
The value of x2−(y−z)2(x+z)2−y2+y2−(x−z)2(x+y)2−z2+z2−(x−y)2(y+z)2−x2. [1 Mark, June 2016]
(a) 0
(b) 1
(c) -1
(d) ∞
Answer:
(b) is correct
Tricks : Cyclic order
Question 19.
If 3x = 5y = (75)z then
(a) 1x+2y=1z
(b) 2x+1y=1z
(c) 1x+1y=1z
(d) None
Answer:
See Short Cut Tricks Book “QUICKER BMLRS”
3x = 5y = (75)z ………(1)
31 × 52 = 751 ………(2)
Tricks:
Power of (2) ÷ power of (1)
and put + sign at the place of “×”
We get 1x+2y=1z
So, (a) is correct
Question 20.
If abc = 2, then the value of 11+a+2b−1+11+b2+c−1+11+a−1+c = [1 Mark, June 2016]
(a) 1
(b) 2
(c) 12
(d) 35
Answer:
TRICKS
“Put a = 1, b = 2 & c = 1. So that abc = 2” in the given question. We get
11+1+22+11+22+1−1+11+1−1+1 = 1
Option (a) is correct.
Question 21.
If a = 6√+5√6√−5√, b = =6√−5√6√+5√ then the value of 1a2+1b2 is [1 Mark, June 2017]
(a) 486
(b) 484
(c) 482
(d) 500
Answer:
Option (a) is correct.
Question 22.
If u5x = v5y = w5z and u2 = vw then xy + zx – 2yz
(a) 0
(b) 1
(c) 2
(d) None of these
Answer:
(a) u5x = v5y = w5z ⇒ ux = vy = wz
Tricks : See Quicker BMLRS Chapter : Indices
u2 = vw;
∴ 2x=1y+1z=y+zyz
or ; xy + zx = 2yz
or; xy + zx -2yz = 0
Question 23.
(a) x-(ap+bq+cr)
(b) xa+b+c
(c) x(ap+bq+cr)
(d) xabc [1 Mark, June 2018]
Answer:
(b)
= xa.xb.xc
= xa+b+c
Option (b) is correct.
Question 24.
2n+2n−12n+1−2n
(a) 12
(b) 32
(c) 23
(d) 13 [1 Mark, May 2018]
Answer:
(b)
Tricks:
Put minimum power = n – 1 = 0 or n = 1 in the question
∴ 2n+2n−12n+1−2n=21+21−121+1−21=2+14−2=32
Question 25.
2m+1×32m−n+3×5n+m+4×62n+m62m+n×10n+1×15m+3 [1 Mark, Nov. 2018]
(a) 32m-2n
(b) 32n-2m
(c) 1
(d) None
Answer:
Tricks
Put m = n = 0 in this equation. 2m+1×32m−n+3×5n+m+4×62n+m62m+n×10n+1×15m+3 = 1
Question 26.
If 2x = 3y2 = 12z2 then [1 Mark, June 2019]
(a) 1×2+1y2=1z2
(b) 1×2+2y2=1z2
(c) 1×2+2y2=1z2
(d) None
Answer:
∵ 2x2 = 3y2 = 12z2 ………..(1) (Given)
Tricks:
Factorize 12 in terms of 2 & 3. We get
22 × 31 = 121 ……….(2)
Always write as power of base of (2) ÷ Power on same base of 1 ; put “+”
Sign at the place of “×” Sign. So;
2×2+1y2=1z2
So (c) is correct.
Details
(c) is correct.
Logarithms
Question 1.
7log(1615) + 5log(2524) + 3log(8180) is equal to
(a) 0
(b) 1
(c) log 2
(d) log 3
Answer:
Calculation Tricks
I Type i6 ÷ 15 × button Then push = button 6 time then push M+ button.
II Type 25 ÷ 24 × button Then push = button 4 times
III Then Push × button Then MRC button 2 time Then Push M+ button
IV Type 81 ÷ 80 × button Then = button 2 times
V Then Push × button and Then MRC button 2 times Then = button
We get; it is approx 2
So. Value = log 2
(c) is Correct
Question 2.
The value of the expression alogab.logbc.logcd.logdt. [1 Mark, Feb. 2007]
(a) t
(b) abcdt
(c) (a + b + c + d + t)
(d) None.
Answer:
alogab.logbc.logcd.logdt
= a1logat = t1 = t
(a) is correct
Question 3.
If log10000x = −14, then x is given by: [1 Mark, Feb. 2007]
(a) 1/100
(b) 1/10
(c) 1/20
(d) None of these
Answer:
log10000x = −14
∴ (10000)−14 = x
or (104)−14 = x
or x = 10-1 = 110
(b) is correct
Question 4.
If log (2a – 3b) = log a – log b, then a = ? [1 Mark, May 2007]
(a) 3b22b−1
(b) 3b2b−1
(c) b22b+1
(d) 3b22b+1
Answer:
log(2a – 3b) = logab
or 2a – 3b = a
or 2ab – 3b2 =a
or 2ab – a = 3b2
or a(2b – 1) = 3b2
or a = 3b22b−1
(a) is correct
Question 5.
1logab(abc)+1logbc(abc)+1logca(abc) is equal to: [1 Mark, Aug. 2007]
(a) 0
(b) 1
(c) 2
(d) -1
Answer:
1logab(abc)+1logbc(abc)+1logca(abc)
= logabc (abJ)c.ca) = logabc (abc)2
= 2 logabc (abc) = 2 × 1 = 2
∴ (c) is correct
Question 6.
Number of digits in the numeral for 264 [Given log 2 = 0.30103] : [1 Mark, Aug. 2007]
(a) 18 digits
(b) 19 digits
(c) 20 digits
(d) 21 digits
Answer:
(c) Let x = 264
log x= 64 log 2
= 64 × 0.30103
= 19.26592
x = AL (19.26592) characteristics = 19
∴ No. of digits in the number = 19 + 1 = 20
∴ (c) is correct
Question 7.
The value log38log916.log410 is: [1 Mark, Nov. 2007]
(a) 3log102
(b) 7log103
(c) 3logez
(d) None
Answer:
(a) is correct
Question 8.
If x = then the value of n is: [1 Mark, Feb. 2008]
(a) 12loge1+x1−x
(b) loge1+x1−x
(c) loge1−x1+x
(d) loge1−x1+x
Answer:
Question 9.
log 144 is equal to:
(a) 2 log 4 + 2 log 2
(b) 4 log 2 + 2 log 3
(c) 3 log 2 + 4 log 3
(d) 3 log 2 -4 log 3
Answer:
log 144 = log(16 × 9)
= log 16 + log 9
= log 24 +log 32
= 4 log 2 + 2 log 3
(b) is correct
Tricks:- Go by choices.
Question 10.
If log2 [log3(log2 x)] = 1, then x equals ; [1 Mark, June 2008]
(a) 128
(b) 256
(c) 512
(d) None
Answer:
log2[log3(log2x)] = 1
Tricks:- 2321 = x
or x = 2321 = 23×3 = 29 = 512
(c) is correct
Detail Method log2 [log3 (log2 x)] = 1
or log3 (log2 x) = 21 = 2
or log2x = 32
or log2 x = 9
or x = 29 = 512
(c) is correct
Question 11.
If log(a+b4)=12 (log a + log b) then: ab+ba
(a) 12
(b) 14
(c) 16
(d) 8
Answer:
log(a+b4)=12 (log a + log b)
or log(a+b4) = log(ab)1/2
or a+b4=ab−−√
or a + b = 4ab−−√
Squaring on both sides; we get (a + b)2 = 16ab
or a2 + b2 + 2ab = 16ab
or a2 + b2 = 14ab
or a2ab+b2ab=14abab [Dividing by ab on both sides]
or ab+ba = 14
(b) is correct
Question 12.
log (m+n) = log m+ log n, m can be expressed as : [1 Mark, June 2009]
(a) m = nn−1
(b) m = nn+1
(c) m = n+1n
(d) m = n+1n−1
Answer:
log(m + n) = log m + log n
or log (m + n) = log(mn)
or m + n = mn
or m – mn = -n
or m (1 – n) = -n
or m = −n1−n=nn−1
∴ (a) is correct
Tricks Go by choices.
Question 13.
log4(x2 + x) – log4(x + 1) = 2. find x. [1 Mark, June 2009]
(a) 16
(b) 0
(c) -1
(d) None of these
Answer:
log4(x2 + x) – log4(x + 1) = 2
or log (x2+xx+1) = 2
or x(x+1)x+1 = 42
or x= 16
∴ (a) is correct
Tricks: Go by choices.
Question 14.
Find the value of [log10 25−−√ – log10(23) + log10(4)2]x: [1 Mark, Dec. 2009]
(a) x
(b) 10
(c) 1
(d) none
Answer:
[log10 25−−√ – log10(23) + log10(4)2]x = [log10(58 × 16)]x
= (log1010)x = 1x = 1
∴ (c) is correct
Question 15.
If loga b + loga c = 0 then: [1 Mark, June 2010]
(a) b = c
(b) b = -c
(c) b = c = 1
(d) b and c are reciprocals.
Answer:
logab + logac = 0
or loga(bc) = loga1
∴ bc = 1
b = 1c
∴ (d) is correct
Question 16.
The value of 2log x + 2 log x2 + 2 log x3 + ………. + 2 log xn will be: [1 Mark, Dec. 2010]
(a) n(n+1)logx2
(b) n(n + 1)log x
(c) n2 log x
(d) None of these
Answer:
Detail Method:
2 log x + 2 log x2 + 2logx3 + ………….. + 2 log xn
=2 log x + 2.2 logx + 2.3logx + ……….. + 2.n.logx
= 2 logx. [1 + 2 + 3 + ………. + n]
= 2 log x. n(n+1)2 = n(n+1)
= (b) is correct
Tricks Put n = 2 in options directly.
This should be equal to sum of 1st 2 terms = 2 logx + 2.2logx = 6 logx
Which gives option (b)
∴ (b) is correct.
Question 17.
Solve: logx10−32+11−logx103 = 2. [1 Mark, Dec. 2010, June 2011]
(a) 10-1
(b) 102
(c) 10
(d) 103
Answer:
logx10−32+11−logx103 = 2
Tricks: Go by choice [Do Mentally]
For (a) x = 10-1
L.H.S = log(10−1)10−32+11−log10−1103
= log(10−1)10−32+11−log10−1103
= -2 + 4 = 2 = (R.H.S)
∴ (a) is correct
Question 18.
If n = m! where (‘m’ is a positive integer > 2) then the value of ∴ 1log2n+1log3n+1log4n+…………+1logmn. [1 Mark, June 2011]
(a) 1
(b) 0
(c) -1
(d) 2
Answer:
Given n = m !
1log2n+1log3n+1log4n+…………+1logmn
= logn2 + logn3 + logn4 + ………… + lognm
= logn(2. 3.4 ……………….. m)
= logn (1 . 2. 3. 4 …………….. m) = log(ml) (ml) = 1
∴ (a) is correct
Question 19.
If log2x + log4x = 6, then the value of x is: [1 Mark, 2011 Dec.]
(a) 16
(b) 32
(c) 64
(d) 128
Answer:
(a) is correct
Tricks: Go by choices for (a) if x = 16
L.H.S = log216 + log4 16 = 4 + 2 = 6(R.H.S)
∴ (a) is correct
Detail Method log2x + log4x = 6
or log2x + log22 x = 6
or log2x + 12log2x = 6
or (1 + 12)log2x = 16
or log2x = 6×23 = 4
∴ x = 24 = 16
Question 20.
If logxY= 100 and log2 x = 10, then the value of‘Y’
(a) 210
(b) 2100
(c) 21,000
(d) 210,000
Answer:
(c) log 2 x = 10 ∴ x = 210
Now logxy = 100 y = ;
c100 y = (210)100 =21000 (c) is correct
Question 21.
Which of the following is true. If 1ab+1bc+1ca=1abc. [1 Mark, Dec.2012]
(a) log (ab + bc +ca) =abc
(b) log (1a+1b+1c) =abc
(c) log (abc) = 0
(d) log(a+b + c) = 0
Answer:
(a) is correct 1ab+1bc+1ca=1abc
Multiplying both sides by abc; abcab+abcbc+abcca=abcabc
or; c + a + b = 1
or; a + b + c= 1
Taking log on both sides ; we get
log (a + b + c) = log 1 = 0
Question 22.
If (logx−−√ 2)2 = logx 2 then x = [1 Mark, June 2013]
(a) 16
(b) 32
(c) 8
(d) 4
Answer:
(a) is correct (logx−−√ 2)2 = logx2
or (logx1/2 2)2 = logx 2
or (112logx2)2 = logx2
or 4(logx2)2 – logx2 = 0
or 4(logx2)2 – logx2 = 0
or logx2[4logx – 1] = 0
If logx2 = 0 (Invalid) 4logx2 – 1 = 0
or 4logx2 = 1
or logx2 = 14
or x1/4 = 2 ⇒ x = 24 = 16
Tricks Go by choices
For (a) LHS = (log^ 2)2 = (log4 2)2
RHS = log16 2 = log24 2 = 14log22 = 14
(a) is correct
Note Never write; check mentally.
Question 23.
Find Value of [logyx.logzy.logx z]3 =
(a) 0
(b) -1
(c) 1
(d) 3
Answer:
(c) is correct [logyx.logzy.logx z]3
= [logxx] = [1]3 = 1
Question 24.
Find the value of Log49.Log32 = [1 Mark, Dec. 2013]
(a) 3
(b) 9
(c) 2
(d) 1
Answer:
(d) is correct Log49.Log32 = l0g(2)2 (32).log32
= 22log23.log32
= 1 × 1 = 1
Question 25.
If X = log2412; y = log36 24; z = log48 36 then xyz + 1 = ? [1 Mark, June 2014]
(a) 2xy
(b) 2zx
(c) 2yz
(d) 2
Answer:
(c) is correct xyz + 1 = log2412. log36 24. log48 36+1
= log4812 + log48 48
= log48 (12 × 48)
= log48(12 × 2)2
= 2log48 24
= 2 log36 24. log48 36
= 2yz
Question 26.
If x2 + y2 = 7xy then log 13(x + y) = [1 Mark, June 2014]
(a) log x + log y
(b) 12(log x + log y)
(c) 13(log x + log y)
(d) 13(log x. log y)
Answer:
Question 27.
If log x= a – b; log y = a + b then log(10xy2)
(a) 1- a – 3b
(b) a – 1 + 3b
(c) a + 3b – 1
(d) 1 – b + 3a
Answer:
(a) is correct
log x = a+b ; log y = a-b.
log(10xy2) = log1010 + log x – log y2
= 1 + a + b – 2log y = 1 + a + b – 2 (a – b)
= 1 + a + b – 2a + 2b = 1 – a + 3b
Question 28.
If x = 1 + logp qr, y = 1 + logq rp and z = 1 + logr pq ; then the value of 1x+1y+1z = ……….
(a) 0
(b) 1
(c) -1
(d) 3
Answer:
(A) is correct 1 1 1
Tricks:- Cyclic order
So, 1x+1y+1z = 1
(See Quicker BMLRS)
Question 29.
If log x = m + n; log y = m – n then log(10xy2) =
(a) 1 – m + 3n
(b) m – 1 + 3n
(c) m + 3n + 1
(d) None
Answer:
(a) If log x = m + n; log y = m – n
Then log(10xy2)
= log 10 + log x – log y2
= 1 + log x – 2 log y
= 1 + (m + n) – 2(m – n)
= 1 + m + n – 2m + 2n
= 1 – m + 3n
∴ (a) is correct.
Question 30.
log3 5. × log5 4 × log2 3: [1 Mark, Dec. 2015]
(a) 2
(b) 5
(c) -2
(d) None of these
Answer:
(a) is correct log3 5. log5 4. log2 3
= log3 4. log2 3
= log2 4 = 2
Question 31.
The integral part of a logarithm is called __________, and the decimal part of a logarithm is called __________. [1 Mark, June 2016]
(a) Mantissa, Characteristic
(b) Characteristic, Mantissa
(c) Whole, Decimal
(d) None of these
Answer:
(b) is correct.
Question 32.
The value of 1log360+1log460+1log560 = __________. [1 Mark, June 2016]
(a) 0
(b) 1
(c) 5
(d) 60
Answer:
(b) is correct. log60 3 + log60 4 + log60 5
= log60(3 × 4 × 5) + log2 60 = 1
Question 33.
If log4(x2 + x) – log4(x + 1) = 2 then the value of x is
(a) 2
(b) 3
(c) 16
(d) 8
Answer:
(c) is correct.
or log4x – 2 ⇒ x = 42 = 16
Question 34.
Given log 2 = 0.3010 and log 3 = 0.4771 then the value of log 24 [1 Mark, Dec. 2016]
(a) 1.3081
(b) 1.1038
(c) 1.3801
(d) 1.8301
Answer:
(c) is correct.
Calculator Tricks:
Type 24 then √ button 19 times – 1 x 227695 = button. We will get the required value of log 24.
Question 35.
log (13 + 23 + 33 + ………. + n3) = .[1 Mark, June 2017]
(a) 2 log n + 2 log (n + 1) – 2 log2
(b) log n + 2 log (n +1)-2 log2
(c) 2 log n + log (n +1)- 2 log 2
(d) None
Answer:
log (13 + 23 + 33 + …………… + n3)
= log(n(n+1)2)2 = 2logn(n+1)2
= 2 [log n + log (n +1) – log 2]
= 2 log n + 2log(n+1) – 2log 2
So, (a) is correct
Tricks: Go by choices
Question 36.
If log3[log4(logxx)] = 0 then X =
(a) 4
(b) 8
(c) 16
(d) 32
Answer:
(c)
Tricks: GBC
for option (c) log3[log4(log2x)]
= log3[log4(log216)]
= log3(log44) = log31 = 0
∴ (c) is correct.
Question 37.
If log(x−y2)=12(log x + log y) then x2 + y2 = _______. [1 Mark, Dec. 2017]
(a) 6 xy
(b) 2xy
(c) 3x2y2
(d) 4x2y2
Answer:
or; x2 + y2 – 2xy = 4xy
or; x2 + y2 =6xy
Question 38.
If log2(3√2) = 115 then x = . [1 Mark, June 2018]
(a) 2
(b) 8
(c) 16
(d) 32
Answer:
(d) log2(3√2) = 115
or x1/15 = 3√2 = 21/15
or x = (21/15)15 = 25 = 32
Question 39.
The value of the expression :
alogab.logb c. logcd. logdt. [1 Mark, Nov. 2018]
(a) t
(b) abcdt
(c) (a + b + c + d +1)
(d) None
Answer:
alogab.logb c. logcd. logdt
= alogat = a1.logat = t1 = t
[Formula axlogab = bx]
Question 40.
log2 log2 log216 = ? [1 Mark, Nov. 2018]
(a) 0
(b) 3
(c) 1
(d) 2
Answer:
(c) log2 log2 log216 = log2 log24 = log22 = 1
Question 41.
The value of
log5(1+15) + log5 (1 + 16) + ……… + log5 (1 + 1624)
(a) 2
(b) 3
(c) 5
(d) 0
Answer:
Question 42.
log22–√ (512): log32–√ 324 = .[1 Mark, June 2019]
(a) 128:61
(b) 2:3
(c) 3 :2
(d) None
Answer:
Calculator Tricks
log22–√ (512) = 5 + 1
Type 2 × 2√ button = button.
Then press x button then continue pressing = button until to get 512
Here = button has been pressed 5 times. So log value
= No. of = button pressings + 1
Similarly For log32–√ 324
Type 3 × 2√ button = button then
x = button 3 times ; we get
log32–√324 value = 3 + 1=4
So; log22–√512:log32–√324
= 6: 4 = 3: 2