Analogy Answer & Solution
Answer: Option B
REASONING Practice Mcq Question And Answer
1.Find the correct option for the mirror image for the following Fig1
Find the correct option for the mirror image for the following Fig1
2.A number when successively divided by 7 and 8 leaves the remainders 3 and 5 respectively. What is the remainder when the same number is divided by 56?
A number when successively divided by 7 and 8 leaves the remainders 3 and 5 respectively. What is the remainder when the same number is divided by 56?
Answer & Solution
Answer: Option A
56 = d1 × d2 ∴ required remainder = d1 r2 + r1,where d1 = 7 and r 1 = 3 and r2 = 5.i.e. 7 × 5 + 3 = 38
3.The value of $3\div \left[ \left( 8-5 \right) \div \left\{ \left( 4-2 \right) +\left( 2+\frac{8}{13} \right) \right\} \right] $
The value of $3\div \left[ \left( 8-5 \right) \div \left\{ \left( 4-2 \right) +\left( 2+\frac{8}{13} \right) \right\} \right] $
Answer & Solution
Answer: Option B
$3\div \left[ \left( 8-5 \right) \div \left\{ \left( 4-2 \right) \div \left( 2+\frac{8}{13} \right) \right\} \right] $$ =3\div \left[ 3\div \left\{ 2\div \frac{34}{13} \right\} \right] $$=3\div \left[ 3\div \frac{13}{17} \right] =3\div \frac{3\times 17}{13}$$=3\times \frac{13}{3\times 17}=\frac{13}{17}$
4.A number when divided by 5 leaves a remainder 3. What is the remainder when the square of the same number is divided by 5?
A number when divided by 5 leaves a remainder 3. What is the remainder when the square of the same number is divided by 5?
Answer & Solution
Answer: Option D
Let the number be 5q + 3, where q is quotientNow, (5q + 3)2 = 25q2 + 30q + 9 = 25q2 + 30q + 5 + 4 = 5[5q2 + 6q + 1] + 4Hence, remainder is 4
5.A number when divided by a divisor, left remainder 23. When twice of the number was divided by the same divisor, remainder was 11. Find the divisor.
A number when divided by a divisor, left remainder 23. When twice of the number was divided by the same divisor, remainder was 11. Find the divisor.
Answer & Solution
Answer: Option D
Let number be N.Then, N = Divisor × Q1 + 232N = Divisor × Q2 + 11,where Q1 and Q2 are quotients respectively.Here, we have two equations and 3 variables. There equations cannot be solved.
6.If ‘n’ is a natural number then the greatest integer less than that or equal to $\left( 2+\sqrt{3} \right) ^{\mathrm{n}}$ is
If ‘n’ is a natural number then the greatest integer less than that or equal to $\left( 2+\sqrt{3} \right) ^{\mathrm{n}}$ is
Answer & Solution
Answer: Option A
putting n = 1, we get $\left( 2+1.\sqrt{3} \right) $ = whose integral part is 3putting n= 2, we get $\left( 2+\sqrt{3} \right) ^2=4+3+4\sqrt{3}$ ; whose integral part is 11 which is again an odd numberNow, through the options it can be judged that the greatest integer must always be an odd number.
7.Two different numbers when divided by the same divisor, left remainder 11 and 21 respectively, and when their sum was divided by the same divisor, remainder was 4. What is the divisor?
Two different numbers when divided by the same divisor, left remainder 11 and 21 respectively, and when their sum was divided by the same divisor, remainder was 4. What is the divisor?
Answer & Solution
Answer: Option B
Divisor = [Sum of remainders] – [ Remainder when sum is divided]= 11 + 21 – 4 = 28
8.How many numbers, lying between 1 and 500, are divisible by 13?
How many numbers, lying between 1 and 500, are divisible by 13?
Answer & Solution
Answer: Option A
Number divisible by 13, 26, 39, .... 494Let n be the total numbers494 = 13 + (n – 1) × 13⇒ 13n = 494⇒ n = 38
9.What is the remainder when 1! + 2! + 3! + …… + 100! Is divided by 7 ?
What is the remainder when 1! + 2! + 3! + …… + 100! Is divided by 7 ?
Answer & Solution
Answer: Option B
7! + 8! + 9! + 10! + ....... + 100 = 7.6! + 8.7.6! + 9.8.7.6! + ....... + 100! is completely divisible by 7 as each of the terms contain at least one 7 in it.Now, 1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873which leaves a remainder of 5 when divided by 7.
10.How many numbers, between 1 and 300 are divisible by 3 and 5 together?
How many numbers, between 1 and 300 are divisible by 3 and 5 together?
Answer & Solution
Answer: Option C
LCM of 3 and 5 = 15Number divisible by 15 are 15, 30, 45 .....300.Let total numbers are n300 = 15 + (n – 1) × 15⇒ 300 = 15 + 15n –15⇒ n = 20