| 1. Solve 299x+301y = 1501; 301x+299y = 1499 | Easy |
| A. x=2, y=3 B. x=2, y=2 C. x=3, y=4 D. None of these |
View Answer
Answer: Option A
Explanation:
Given, 299x+301y = 1501 —– ❶
301x+299y = 1499 —– ❷
Adding ❶ & ❷ -> 600(x+y) = 3000 -> x+y = 5 —– ❸
Subtract ❷ from ❶
2(y-x) = 2 -> y-x = 1 —– ❹
Solving ❸&❹ x = 2, y = 3
Alternatively
Plug in values from the options in the given equations.
| 2. Solve (x/7)+ y = 20/7; 13/(x+(y/4)) = 2 | Easy |
| A. x=6, y=3 B. x=-2, y=6 C. x=6, y=2 D. None of these |
View Answer
Answer: Option C
Explanation:
x/7 + y = 20/7 -> x+7y = 20 —– ❶
Now, 13/(x+y/4) = 2
-> (13/1)/(4x/4+y/4) = (13*4)/(4x+y) = 2
-> 52 = 2(4x+y)
-> 4x+y =26 —– ❷
Solve ❶ & ❷ simultaneously to get x = 6, y = 2
Alternatively
Plug in values from the options in the given equations.
| 3. Solve 5/(x+2y) + 4/(2x+y) = 2; 10/(x+2y) – 1/(2(2x+y)) = 15/8 | Medium |
| A. x=2, y=1 B. x=1, y=-1 C. x=-2, y=-1 D. None of these |
View Answer
Answer: Option D
Explanation:
Let 1/(x+2y) = a and 1/(2x+y) = b
then,
5a+4b = 2 —– ❶
10a – b/2 = 15/8 or 80a-4b = 15 —– ❷
Solving simultaneously a = 1/5, b = 1/4
-> x+2y = 5 and 2x+y = 4
Solving simultaneously, x = 1, y = 2
Alternatively
Plug in values from the options in the given equations.
| 4. Four years ago, a father was four times the age of his son. Four years hence, the father will be 2.8 times the age of his son. What is the present age of the father? | Easy |
| A. 48 B. 52 C. 56 D. None of these |
View Answer
Answer: Option B
Explanation:
Let son’s present age be x.
-> 4(x-4) + 56 = 2.8 (x+4)
-> x = 16
-> Father was 4(16-4) = 48 years 4 years ago
-> Father’s current age = 52 years
| 5. A fraction becomes ½ if its numerator is increased by 5 and the denominator by 7. It becomes 3/5 if the numerator is increased by 19 and denominator by 28. Find the fraction. | Easy |
| A. 2/7 B. 4/21 C. 1/8 D. None of these |
View Answer
Answer: Option A
Explanation:
Let the fraction be x/y
(x+5)/(y+7) = 1/2 or 2x+10 = y+7 -> 2x-y = -3 —– ❶
(x+19)/(y+28) = 3/5 or 5x+95 = 3y+84 -> 5x-3y = -11 —– ❷
Solving ❶&❷ simultaneously, x= 2, y =7, fraction = 2/7
Alternatively
Plug in values from the options.
| 6. A two-digit number is such that the sum of its digits is two times the difference of its digits. Find the number if the number exceeds the number formed by reversing the digits by 36. | Easy |
| A. 31 B. 62 C. 93 D. 26 |
View Answer
Answer: Option B
Explanation:
Let the units digit of the number be y and the ten’s digit be x
-> the number can be written as (10x + y) and the number formed by reversing the digits can be written as (10y + x).
Now, x+y = 2(x-y) —– (x>y, because 10x+y> 10y+x)
x+y = 2x-2y ->x = 3y
x = 3y —– ❶
Also, (10x+y) -(10y+x) = 36
10x+y-10y-x = 36 -> 9x-9y = 36
-> x-y = 4 —– ❷
From ❶ & ❷, y = 2, x = 6
Thus, the number is (10*6)+2 = 62
Alternatively
Apply the given conditions to each of the options to get option B as the answer.
| 7. There is some money with A and some with B. If A gives Rs. 20 to B, then the amounts with them would be interchanged. If instead B gives Rs. 40 to A, then A would have Rs. 100 more than that of B. Find the amount that A and B initially have. | Medium |
| A. 100, 80 B. 200, 180 C. 300, 280 D. Can’t be determined |
View Answer
Answer: Option D
Explanation:
Let the amount with A be x, and that with B be y.
-> x-20 = y and y+20 = x
-> x – y =20 —❶
Also, x+40 = y-40+100
-> x-y = 20 —-❷
Unique solution not possible as both equations are the same.
First three options meet all the conditions; hence the risk of approaching the question through options, as one may settle for the first option which satisfies the given conditions.
| 8. Alexa started playing a money game. In the first round, she quadrupled her amount, and he gave away a certain amount y to her friend Monica. In the second round she tripled the amount with her and gave away an amount 2y to Monica. In the third round she doubled the amount with her and gave away an amount 3y to Monica. In the fourth round she again doubled the amount with her and gave away an amount 4y to Monica, and was finally left with no money. If she gave away a total of Rs. 960 to Monica, then what was the money she had at the end of the third round? | Difficult |
| A. Rs. 192 B. Rs. 60 C. Rs. 240 D. None of these |
View Answer
Answer: Option A
Explanation:
Let the amount with Alexa be x rupees. Then:
| Round | Amount with Alexa | Gave | Left |
| 1 | 4x | y | 4x-y |
| 2 | 12x-3y | 2y | 12x-5y |
| 3 | 24x-10y | 3y | 24x-13y |
| 4 | 48x-26y | 4y | 48x-30y |
Therefore, 48x-30y = 0 or x = 5y/8 —– ❶
Now, y+2y+3y+4y = 960 -> y = 96 —– ❷
So, from ❶ & ❷, x = (5*96)/8 = 60
At the end of third round, Alex had an amount 24x-13y = (24*60 – 13* 96) = 192
| 9. 100 girls are standing in a line, not all of whom have the same number of pencils with them. If the first girl distributes her pencils to the remaining ninety-nine girls, quadrupling the number of their pencils, she will be left with three pencils. If the hundredth girl takes away one pencil from each of the remaining ninety-nine, then three times the number of pencils she will have will be twenty-one less than the number of pencils the first girl initially had. What is the total number of pencils with the second girl to the ninety-ninth girl? | Difficult |
| A. 99 B. 105 C. 297 D. Can’t be determined |
View Answer
Answer: Option B
Explanation:
Let the number of pencils with the first girl be x, total number of pencils with the second girl to the ninety-ninth girl be y, and number of pencils with the hundredth girl be z.
The first girl distributes in a way so as to quadruple the number of pencils with the other 99 girls. The other 99 girls had (y+z) pencils initially; so, if their number is quadrupled, it implies that the first girl gave away 3(y+z) pencils and is eventually left with only 3 pencils.
-> x-3(y+z) = 3 or x-3y-3z = 3—– ❶
Also, 3(z+99) = x-21 -> x-3z = 318 —– ❷
From ❶ & ❷ y = 105
| 10. The difference between a three-digit number and the number formed by reversing the digits is 99. The sum of the units and tens digit is the same as the difference of the hundreds and the units digit. Also, the hundreds digit is twice the unit’s digit. Find the number. | Medium |
| A. 998 B. 301 C. 201 D. None of these |
View Answer
Answer: Option C
Explanation:
Let the number be abc i.e., 100a+10b+c
-> (100a+10b+c)-(100c+10b+a) = 99
-> a-c = 1 —– ❶
Also, b+c = a-c
-> b+c = 1 —– ❷
Also, a = 2c —– ❸
From ❶,❷&❸
a = 2, b = 0, c = 1
Therefore, the number is 201.