Q.1: If a : b = 3 : 4, b : c = 2 : 3, c : d = 1 : 2, find the value of ((a+b+c+d)/(b+d))^½
A) 5:4
B) 4:5
C) 1:1
D) 2:3
Explanation: Given a:b=3:4 we can write: a=3k, b=4k Given b:c=2:3 knowing b=4k 4k/c=2/3 ⟹ c=(3/2)*4k= 6k Given c:d=1:2 and knowing c=6k 6k /d =1/ 2 ⟹ d=2*6k = 12k ,a+b+c+d=3k+4k+6k+12k = 25k ,b+d=4k+12k = 16k ,a+b+c+d = 16k/25k =16/25,(a+b+c+d)/(b+d) =16k/25k =16/25 √((a+b+c+d)/(b+d)) = √(16/25) = 4/5 answer B
Q.2: Rupees 812.5 is divided among Subhas, Ragini, and Gautam in such a way that 3-times Subhas’s share, 2-times Ragini’s share and 4 times Gautam’s share is equal. Calculate their individual share.
A) 10
B) 12
C) 13
D) 21
Explanation: A six digit number in the form of xyzxyz is always divisible by 1001 i.e. 7*11*13
5a4b6c a=6, b=5, c=4
3a-5b+7c
3*6-5*5+7*4=21
5a4b6c a=6, b=5, c=4
3a-5b+7c
3*6-5*5+7*4=21
Q.3: The number A4531B is divisible by 72. Where A and B single digit number then the number of factors of A+B is
A) 5
B) 9
C) 2
D) Can’t determine
Explanation: A number is also divisible by divisor’s co factors. A4531B is divisible by 72 (9*8) so it should be divisible by 9 and 8 also.
To be divisible by 8, the last 3 digits (31B) should be divisible by 8 therefore B = 2
A number is divisible by 9 when sum of all digits is divisible by 9
A+4+5+3+1+2 = A + 15
so A =3 A+B = 5 factors of 5 = 5*1 Ans is 2 .
To be divisible by 8, the last 3 digits (31B) should be divisible by 8 therefore B = 2
A number is divisible by 9 when sum of all digits is divisible by 9
A+4+5+3+1+2 = A + 15
so A =3 A+B = 5 factors of 5 = 5*1 Ans is 2 .
Q.4: When (224 – 1) is divided by 7 , the remainder is:?
A) 4
B) 2
C) 1
D) 0
Explanation: (23)8 -1/7 = 88 – 1/7 By remainder rule 8/7 rem. Is 1 so 1-1 is 0
.
Q.5: A number is divided by 406 leaves remainder 115. What will be the remainder when it is divided by 29??
A) 31
B) 28
C) 27
D) can’t determine
Explanation: The required number N= 406*Q+115 by division theorem
when 406 /29 remainder=0 and 115/29 remainder= 28 so ans is 28 .
when 406 /29 remainder=0 and 115/29 remainder= 28 so ans is 28 .
Q.6: If the 9-digit number 807x6y9z8 is divisible by 99, then the value of (x + y + z)^½ is
A) 4
B) 3√3
C) 6
D) 3√5
Explanation:given number is divisible by 99 if it is divisible by 9 and 11 both (99=11*9). check divisibility by 9 if the sum of its digits is divisible by 9. sum of the digits: 8+0+7+x+6+y+9+z+8 = x+y+z +38 is divisible by 9 if x+y+z = 7, 16
Now check Check divisibility by 11 : A number is divisible by 11 if the alternating sum of its digits is divisible by 11. alternating sum of the digits: 8−0+7−x+6−y+9−z+8 = 38 – x–y–z is divisible by 11 if x+y+z = 16 So x + y + z = 16 (x + y + z)^½ = 4
Now check Check divisibility by 11 : A number is divisible by 11 if the alternating sum of its digits is divisible by 11. alternating sum of the digits: 8−0+7−x+6−y+9−z+8 = 38 – x–y–z is divisible by 11 if x+y+z = 16 So x + y + z = 16 (x + y + z)^½ = 4
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