| All 3 digit numbers between 200 and 300 that have a digit which is the average of other 2 digits | 11 |
-
Solution
All numbers between 200 and 300 have digit 2 in the hundred's place.
Now consider the following cases:-
(A)Average - 1, Numbers possible 201, 210
(8) Average - 2, Numbers possible 204, 240, 213, 231, 222
(C)Average - 3, Numbers possible 234, 243
(D) Average - 4, Number possible 246, 264
(E) Average - 5, Numbers possible 258,285
It is not possible to have 6 or a higher digit as the average of other 2 digits.
∴ The numbers possible are 2 + 5 + 2 + 2 + 2 = 13
| how many seventeenths of an inch are 1 centimeter | 5.85 |
-
Solution
2.55 em = 1 inch.
∴ Seventeenth of an inch = \(\frac{2.55}{17}\) = 0.15 cm.
∴ 1 cm = \(\frac{1}{0.15}\) = 6.66 seventeenths of an inch.
∴ Col. A = 6.66, Col. B = 5.85
| The least amount, Maria can spend so that having the furniture shipping cost will not be more expensive than paving sales tax: |
3330 |
-
Solution
In order to save money by having the furniture shipped, sales tax on her purchase must be greater than fixed shipping fee of $429.
∴ In the limiting condition, 13% of purchase amount = $429.
∴ Minimum purchase amount
=\(\frac{429\times 100}{13}=\$ 3300\)
∴ Col. A = $3300, Col. B = $3330
| \(\left | m^{2}-n^{2} \right |\) | m2 – p2 |
-
Solution
m, n, p can take positive as well as negative values.
Let m = 1, n = 2, P = 3
∴ Col. A = 11 - 4 I = 3
Col. B = 1 - 9 = -8
∴ Col. A > Col. B
Let m = -3, n = -2, P = -1
∴ Col. A= I q - 41 = 5
Col. B = 9 - 1 = 8
∴ Col. B > Col. A
| \(\frac{Area\, of\, Sq.B}{Area\, of\, Sq. A}\) | \(\frac{14}{3}\times \left [\frac{perimeter\, of\, Sq.B}{perimeter\, of\, Sq. A} \right ]\) |
-
Solution
If the perimeter of a square is 4 times, the area has to be 16 times.
Area of square A = 6 cm2
Area of square B = 96 cm2
∴ Col. A = 96 = 16
∴ Col. B = \(\frac{14}{3}\) × 4 =\(\frac{56}{3}\) > 16
| \(\frac{p+q}{q+r}\) | \(-\left |\frac{p+q}{q+r} \right |\) |
-
Solution
It is given that p, q, r are all negative numbers. Addition of 2 negative numbers always negative.
∴ p + q and q + r, both have negative values.
∴ Col. A is positive.
Col. B is negative as \(\left |\frac{p+q}{q+r} \right |\) is always positive.
∴ Col. A > Col. B
Note: The problem can also be solved by plugging numbers for p, q, r. E.g. p = -3, q = -2, r =-1.
| Remainder when N is divided b 2 | Remainder when N is divided b 3 |
-
Solution
The sum of three consecutive positive integers is always divisible by 3.
∴ Col. B = 0
However, whether the sun of 3 consecutive integers is even or odd depends upon the choice of numbers.
E.g. if numbers are 1, 2, 3, Col. A = 0
If numbers are 2, 3, 4, Col. A = 1
| markes in the fifth test | 68 |
-
Solution
Total -marks obtained in the first four tests is 58 × 4 = 232.
Total marks obtained in 2nd, 3rd, 4th, 5th tests = 60 × 4 = 240.
∴ Difference between the marks obtained in 5th and 1st test = 240 - 32 = 8
Let marks obtained in first test be 7x and that in 5th be 8x.
∴ 8x - 7x = 8 ., x = 8
∴ Marks in fifth subject = 8x = 64
| If the same sum is lent at simple interest at the same rate, number of years it would take for the sum to be 3 times | 10 |
-
Solution
Principal of $100 becomes $121 after 2 years, therefore the rate of interest must be 10%.
∴ Principle of $100 will become 3 times i.e. $300 after 20 years.
∴ Col. A = 20, Col. B = 10
| cost price of a cap in $ | 60 |
-
Solution
Cost Price of 19 caps - Selling Price of 19 caps = Cost Price of 7 caps
∴ Cost Price of 12 caps = Selling Price of 19 caps = $720
∴ Cost Price of 1 cap = \(\frac{720}{12}\) = 60
Col. A = Col. B