| 1. The ratio between the marks scored by A and B in an exam is 5:4. If A scored 20 marks more than B, find the marks scored by A. | Easy |
| A. 80 B. 100 C. 60 D. None of these |
View Answer
Answer: Option B
Explanation:
A/B = 5/4 —– ❶
A = 20+B —– ❷
-> (20+B)/B = 5/4
-> 80 + 4B = 5B
-> B = 80
-> A = 100
| 2. The ratio between two numbers is 5:6. What number must be added to them so that the ratio between these numbers becomes 7:8 and the sum of their terms becomes 60? | Easy |
| A. 4 B. 8 C. 12 D. None of these |
View Answer
Answer: Option B
Explanation:
Let the numbers be 5x and 6x, let y be added to these numbers
-> (5x+y)/(6x+y) = 7/8
-> y = 2x —– ❶
Also (5x+y) + (6x+y) = 60
-> 11x+2y = 60 —– ❷
From ❶ & ❷, 11x+4x = 60
x = 4, and y = 8
| 3. If a: b = 3: 4, b: c = 5: 6, and c: d = 2: 3, find a: b: c: d. | Easy |
| A. 15:20:14:36 B. 15:20:24:32 C. 15:20:24:36 D. None of these |
View Answer
Answer: Option C
Explanation:
The common part between a: b and b: c is b. So, we make b the same in both ratios to get values of a and c. LCM of 4 and 5 is 20.
Given, a: b = 3: 4 -> a: b = 15: 20 —– ❶
Also, given b: c = 5: 6 -> b: c = 20: 24 —– ❷
From ❶ & ❷, a: b: c = 15: 20: 24
Now c: d = 2: 3 -> when c is 24, d is (3/2)*24 = 36
-> a: b: c: d = 15: 20: 24: 36
Alternatively
Check options to conclude that option C is the answer.
| 4. If a: b = 5: 6, determine (5a+4b)/(4a+3b). | Easy |
| A. 49/38 B. 31/22 C. 31/28 D. None of these |
View Answer
Answer: Option A
Explanation:
Dividing both the numerator & denominator by b, we get
(5a+34)/(4a+3b) = (5a/(b+4))/(4a/(b+3))
Putting a/b = 5/6, we get ((5*5)/6+4)/((4*5)/6+3) = 49/38
| 5. The ratio between the incomes of A and B is 5:4 and the ratio between their expenditures is 4:3. If A saves one-fourth of his income, find the ratio between the savings of A & B. | Medium |
| A. 11/19 B. 12/19 C. 20/19 D. None of these |
View Answer
Answer: Option C
Explanation:
| Income | Expenditure | Savings (S) | |
| A | 5x | 4y | 5x – 4y |
| B | 4x | 3y | 4x -3y |
For A, 5x – 4y = 5x/4
-> 20x – 16y = 5x
-> x/y = 16/15 —– ❶
Therefore, desired ratio = (5x-4y)/(4x-3y)
Dividing numerator and denominator by y, we get
(5x/y – 4)/(4x/y – 3) —– ❷
Putting x/y = 16/15 in ❷, we get As/Bs = 20/19
| 6. The ratio between the ages of mother and son is 8: 3. Mother was 28 years old when she gave birth to her only daughter. If son is 8 years elder to the daughter, find the present age of the mother. | Medium |
| A. 24 B. 32 C. 36 D. None of these |
View Answer
Answer: Option B
Explanation:
Let mother’s age be 8x, then son’s age will be 3x
-> daughter’s age = 3x-8
Now when daughter was born, mother’s age was 8x-(3x-8) = 28 (given) -> x = 4
Hence, mother’s present age = 8x = 8*4 = 32 years
| 7. If a+b: b+c: c+a = 5: 6: 7 and a+b+c = 18. What is the value of abc/4? | Easy |
| A. 32 B. 24 C. 48 D. None of these |
View Answer
Answer: Option C
Explanation:
Let k be a constant. Then,
a+b = 5k —– ❶
b+c = 6k —– ❷
c+a = 7k —– ❸
Adding ❶, ❷ & ❸, we get 2(a+b+c) = 18k
-> a+b+c = 9k
-> k = 18/9 = 2
Now, a = (a+b+c) – (b +c) = 18-(6*2) = 6
Similarly, b =4, and c= 8
-> abc/4 = (6*4*8)/4 = 48
| 8. A, B, C and D bought a product for Rs. 90000. A paid one fifth of the sum of the amounts paid by the others, B paid two third of the sum of the amounts paid by the others, and C paid one third of the sum of the amounts paid by the others. How much did D have to pay? | Medium |
| A. Rs. 36000 B. Rs. 22500 C. Rs. 16500 D. None of these |
View Answer
Answer: Option C
Explanation:
A: others = 1: 5
-> A paid= (1/6) * 90000 = 15000
B: others = 2:3
-> B paid = (2/5) * 90000 = 36000
C: others = 1:3
-> C paid = (1/4) * 90000 = 22500
-> D paid = 90000 – 15000 – 36000 – 22500 = 16500
| 9. P is twice as old as Q is today. A few years back, he was thrice as old and half that number of years prior to that, he was 5 times as old, and 6 years before that he was as old as Q is today. What is P’s age, and how many years back was P thrice as old as Q? | Difficult |
| A. 48, 12 B. 48, 6 C. 56, 12 D. None of these |
View Answer
Answer: Option A
Explanation:
Let Q’s age be x, then P’s age will be 2x
Let the few years be y years.
-> 2x-y = 3(x-y)
-> x = 2y —– ❶
Also, it is given that 2x-y –(y/2) = 5(x-y-(y/2))
-> x = 2y (same as ❶)
Further, 2x-y-(y/2)-6 = x
-> x = (3y/2) + 6 —– ❷
From ❶ & ❷, we get x = 24
P’s age = 2*24 = 48
Also, from ❶, y = x/2 = 12
| 10. The ratio of a two-digit natural number to a number formed by reversing its digits is 13: 31. Which of the following is the sum of the differences of all the numbers of all such pairs? | Difficult |
| A. 108 B. 264 C. 216 D. None of these |
View Answer
Answer: Option A
Explanation:
Let the two-digit number be ab, which in the notational form can be written as a+10b. Number formed by reversing the digits is ba, which can be written as 10b+a.
Given, (10a+b)/(10b+a) = 13/31
-> 310a+ 31b = 130b + 13a
-> 297a = 99b
-> a/b = 1/3
Two-digit numbers satisfying this condition are 13, 26 and 39. The numbers formed by reversing the digits are 31, 62 and 93.
Required sum = (31-13)+(62-26)+(93-39)
= 108