Arithmetic Reasoning Questions and Answers Pdf
Question: 1
A man has a certain number of small boxes to pack into parcels. If
he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs
7 in a parcel, none is left over. What is the number of boxes, he may
have to pack?
(A) 300
(B) 301
(C) 303
(D) 309
Ans: B
Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4,
5, 6 and no remainder when divided by 7.
Thus, the number must be of the form (L.C.M. of 3, 4, 5, 6) x + 1 i.e. (60x + 1) and a multiply of
7. Clearly, for x = 5, the number is multiply of 7. So, the number is 301.
Question: 2
A student got twice as many sums wrong as he got right. If he
attempted 48 sums in all, how many did he solve correctly?
(A) 12
(B) 14
(C) 16
(D) 18
Ans: C
Suppose the boy got x sums right and 2x sums wrong. Then,
x + 2x = 48
Question: 1
A man has a certain number of small boxes to pack into parcels. If
he packs 3, 4, 5 or 6 in a parcel, he is left with one over; if he packs
7 in a parcel, none is left over. What is the number of boxes, he may
have to pack?
(A) 300
(B) 301
(C) 303
(D) 309
Ans: B
Clearly, the required number would be such that it leaves a remainder of 1 when divided by 3, 4,
5, 6 and no remainder when divided by 7.
Thus, the number must be of the form (L.C.M. of 3, 4, 5, 6) x + 1 i.e. (60x + 1) and a multiply of
7. Clearly, for x = 5, the number is multiply of 7. So, the number is 301.
Question: 2
A student got twice as many sums wrong as he got right. If he
attempted 48 sums in all, how many did he solve correctly?
(A) 12
(B) 14
(C) 16
(D) 18
Ans: C
Suppose the boy got x sums right and 2x sums wrong. Then,
x + 2x = 48
