SSC CGL 20191)Let a, b and c be the fractions such that a < b < c. If c is divided by a, the result is \({5\over2}\), which exceeds b by \({7\over4}\).
If a + b + c = 1\({11 \over 12}\), then (c - a) will be equal to :
\({1 \over 2}\)
ATQ,
\(\frac{c}{a} = \frac{5}{2};\)
c = \(\frac{5a}{2};\)
b = \(\frac{5}{2} - \frac{7}{4} = \frac{3}{4};\)
a + b + c = \(1\frac{11}{12} = \frac{23}{12};\)
a +\( \frac{3}{4} + \frac{5a}{2} = \frac{23}{12};\)
\(\frac{7a}{2} = \frac{23}{12} - \frac{3}{4};\)
7a = \( \frac{23}{6} - \frac{3}{2};\)
7a = \( \frac{7}{3} \);
a = \( \frac{1}{3} \);
c =\( \frac{5}{2} \times \frac{1}{3} = \frac{5}{6};\)
c - a =\( \frac{5}{6} - \frac{1}{3} = \frac{1}{2}\)
SSC CGL 20192)If a nine-digit number 389x 6378y is divisible by 72, then the value of \(\sqrt{6x+7y}\) will be :
8
389x6378y is divisible by 72,
Factor of 72 = \(8 \times 9;\)
So, number is divisible by 8 and 9 both.
Divisibility rule for 8,
78y (last three digits should be divisible by 8);
784 is divisible by 8 so,
Value of y = 4;
Divisibility rule of 9,
3 + 8 + 9 + x + 6 + 3 + 7+ 8 + 4
= 48 + x;
54 is divisible by 9;
So, x = 54 - 48 = 6;
Value of\( \sqrt{6x + 7y} = \sqrt{6 \times 6 + 7 \times 4} = \sqrt{36 + 28} =\sqrt{64} = 8\)
SSC CGL 20193)When 12, 16, 18, 20 and 25 divide the least number x, the remainder in each case is 4 but x is divisible by 7. What is the digit at the thousand's place in x ?
8
Number = (LCM of 12, 16, 18, 20 and 25)k + 4
= 3600k + 4;
The number should be divisible by the 7 so,
Value of K = 5;
Number = \(3600 \times 5 + 4 \)= 18000 + 4 = 18004;
The digit at the thousands’ place = 8
SSC CGL 20194)When 7897, 8110 and 8536 are divided by the greatest number x, then the remainder in each case is the same. The sum of the digits of x is:
6
Let the remainder be k. 7897 - k = ax ; 8110 - k = bx; 8536 - k = cx ; Common factor is x. So difference between the numbers, 8110 - 7897 = 213 ; 8536 - 8110 = 426; 8536 - 7897 = 639 ; HCF of 213, 426 and 639 is 213. x = 213; Sum of the digits of x = 2 + 1 + 3 = 6
SSC CGL 20195)One of the factors of\( (8^{2k} + 5^{2k})\), where k is an odd number, is:
89
\((8^{2k} + 5^{2k})\),k is odd nuber so,
Let the k be 1.
=\((8^{2} + 5^{2})\)
= 64 + 25 = 89
SSC CGL 20196)Let \(x= (633)^{24} - \)\((277)^{38} + (266)^{54}\). What is the units digit of x ?
8
x = \((633)^{24} - (277)^{38} + (266)^{54}\)For the unit digit,
24 = 4 \times 6 + 0(remainder);
38 = 4 \times 9 + 2(remainder);
54 = 4 \times 13 + 2(remainder);
Now,
(Base number unit digit)^{remainder}
= \((3)^0 - (7)^2 + (6)^2\);
On consider unit digit,
= 1 - 9 + 6 = 7 - 9;
or 17 - 9 = 8;
8 is the units digit of x.
SSC CGL 20197)The sum of the digits of a two-digit number is \( \frac{1}{7} \) of the number. The units digit is 4 less than the tens digit. If the number obtained on reversing its digits is divided by 7, the remainder will be:
6
Let the number be (10a + b).
ATQ,
a + b =\( \frac{10a + b}{7}\);
7a + 7b = 10a + b;
6b = 3a;
2b = a ---(1);
a - b = 4 ---(2);
From eq (1) and (2),
2b - b = 4;
b = 4;
a = \(4 \times 2
\)= 8;
Number = 10a + b = \(10 \times 8 + 4 =\) 84;
reverse of the number = 48;
Remainder after divide by 7 = 48/7 = 6
SSC CGL 20198)If x is the remainder when \(3^{61284}\) is divided by 5 and y is the remainder when \(4^{96}\) is divided by 6, then what is the value of \((2x-y)\) ?
-2
x is the remainder when \(3^{61284}\) is divided by 5; So, \({3^{61284}\over5}={3^{4\times15321}\over5}\) ; \({3^4\over5}={81\over5}\); Remainder = 1; x=1;
y is the remainder when \(4^{96}\) is divided by 6; remainder is always '4'; thereforey = 4.
2x-y = 2-4 = -2
SSC CGL 20199)The LCM of two numbers x and y is 204 times their HCF. If their HCF is 12 and the difference between the numbers is 60, then x + y = ?
348
SSC CGL 201910)In finding the HCF of two numbers by division method, the last divisor is 17 and the quotients are 1. 11 and 2, respectively. What is sum of the two numbers?
816