Practise Tests
31
Nikhil’s mother asks him to buy 100 pieces of sweets worth 100/-. The sweet shop has 3 kinds of sweets, kajubarfi, gulabjamun and sandesh. Kajubarfi costs 10/- per piece, gulabjamun costs 3/- per piece and sandesh costs 50 paise per piece. If Nikhil decides to buy at least one sweet of each type, how many gulabjamuns should he buy? [XAT2012]
Answer : Option C
Explanation:
Check the key from XAT 2012
32
Three truck drivers, Amar, Akbar and Anthony stop at a road side eating joint. Amar orders 10 rotis, 4 plates of tadka, and a cup of tea. Akbar orders 7 rotis, 3 plates of tadka, and a cup of tea. Amar pays 80 for the meal and Akbar pays 60. Meanwhile, Anthony orders 5 rotis, 5 plates of tadka and 5 cups of tea. How much (in ) will Anthony pay? (XAT 2012) Qno 77
Answer : Option D
Explanation:
Catch the solution in XAT 2012 paper
33
Of 128 boxes of oranges, each box contains at least 120 and at most 144 oranges. The number of boxes containing the same number of oranges is at least
Answer : Option
Explanation:
CAT 2003 solution
34
Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took 1/3 of the mints, but returned four because she had a momentary pang of guilt. Fatima then took 1/4 of what was left but returned three for similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?
Answer : Option B
Explanation:
CAT 2003 solution
35
When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed?
Answer : Option B
Explanation:
Let 10x + y be a two digit number, where x and y are positive single digit integers and x > 0. Its reverse = 10y + x Now, 10y + x – 10x – y = 18 ∴ 9(y – x) = 18 ∴ y – x = 2 Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9) ∴ Other than 13, there are 6 such numbers.
36
Consider all four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?
Answer : Option A
Explanation:
Let aabb (a ≠ 0, a and b being single digits) be a perfect square.
aabb = 1100a + 11b = 11(100a + b)
Also, as aabb is a perfect square, it is a multiple of 121.
∴ aabb = 121K, where K is also perfect square. For K = 4, aabb is a 3 digit number, while for K > 81, K is a 5 digit number.
For 81 ≥ K ≥ 9,
121 × 9 = 1089
121 × 16 = 1936
121 × 25 = 3025
121 × 36 = 4356
121 × 49 = 5929
121 × 64 = 7744
121 × 81 = 9801
∴ There is only one number 7744 of the form aabb, which is a perfect square. Hence, option 1
37
Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?
Answer : Option C
Explanation:
The total number of ways is 18
38
A confused bank teller transposed the rupees and paise when he cashed a cheque for Shailaja, giving her rupees instead of paise and paise instead of rupees. After buying a toffee for 50 paise, Shailaja noticed that she was left with exactly three times as much as the amount on the cheque. Which of the following is a valid statement about the cheque amount?
Answer : Option D
Explanation:
Let the amount on Shailaja’s cheque be Rs. x and paise y = (100x + y) paise (x and y are positive integers)
The teller gives her (100y + x) paise.
Now, 100y + x – 50 = 3(100x + y)
∴ 97y – 299x = 50
∴ y = (50 + 299x)/97 = [50 + 8x + 291x]/97 = [(50 + 8x)/97] + 3
Now as y is an integer,
(50 + 8x) has to be a multiple of 97 with x, y ≤ 99
50 + 8x = 97k (k is an integer)
∴ x = 12k – 6 + [(k – 2)/8] ∴ k = 2, 10, 18…
∴ x = 18, 115, 212…
∴ x = 18 is the only possible value.
This implies that y = 5
∴ The amount on Shailaja’s cheque is over Rs. 18 but less than Rs. 19.
39
A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?
Answer : Option D
Explanation:
Let there be n rows and a students in the first row.
∴ Number of students in the second row = a + 3
∴ Number of students in the third row = a + 6 and so on.
∴ The number of students in each row forms an arithmetic progression with common difference = 3
The total number of students = The sum of all terms in the arithmetic progression 
Now consider each options:
In case of option(4), n = 6 is not possible.
40
If x = (), then x divided by 70 leaves a remainder of
Answer : Option A
Explanation:
X = ()
+
35 × (an odd number) + 35 × (another odd number)
35 × (an even number) = 35 × (2k) … (k is a positive integer)
∴ X = 70 k
∴ X is divisible by 70 (as it is multiple of 70)
Remainder when X is divided by 70 is 0. Hence, option 1.
