Answer: c.
Explanation:
The sum of the consecutive odd numbers must be divisible by 10, according to the sum. Let the numbers be a, a + 2, a + 4, a + 6. Then sum = 4a + 12 = 4(a + 3). Trying different multiples of 40, we get 4(a + 3) = 160, or a = 37. Hence the numbers are 37, 39, 41, 43.
Explanation:
The sum of the consecutive odd numbers must be divisible by 10, according to the sum. Let the numbers be a, a + 2, a + 4, a + 6. Then sum = 4a + 12 = 4(a + 3). Trying different multiples of 40, we get 4(a + 3) = 160, or a = 37. Hence the numbers are 37, 39, 41, 43.
2.
Answer: c.
Explanation:
A covers 60 km in 2 hours. So B takes 60/10 = 6 hrs to meet A. In 6 + 2 hrs, A would have covered 30 x 8 = 240 km. So K takes 240/60 = 4 hrs to overtake A. Hence K could start 8 - 4 = 4 hrs after A.
Explanation:
A covers 60 km in 2 hours. So B takes 60/10 = 6 hrs to meet A. In 6 + 2 hrs, A would have covered 30 x 8 = 240 km. So K takes 240/60 = 4 hrs to overtake A. Hence K could start 8 - 4 = 4 hrs after A.
3.
Correct Answer (B)
Solution:
Express the number 120 as a product of powers of prime factors.
In this case, 120 = (2*2*2) * 3 * 5.
The three prime factors are 2, 3 and 5.
To find the number of factors / integral divisors that 120 has, increment the powers of the prime factors by 1 and then multiply them.
In this case, (3+1) * (1 + 1) * (1 + 1) = 4 * 2 *2 = 16.
4.
5.
6.
Solution:
Express the number 120 as a product of powers of prime factors.
In this case, 120 = (2*2*2) * 3 * 5.
The three prime factors are 2, 3 and 5.
To find the number of factors / integral divisors that 120 has, increment the powers of the prime factors by 1 and then multiply them.
In this case, (3+1) * (1 + 1) * (1 + 1) = 4 * 2 *2 = 16.
Now, the sum of the divisors = [(23+1 - 1)(31+1 - 1)(51+1 -1)]/[(2-1)(3-1)(5-1)] = 360
Note:
If a number n can be expressed as n = ap * bq * cr, where a, b, c are primes, then, No. of divisors (factors) of n = (p+1)(q+1)(r+1).
And the sum of the divisors = [(ap+1 - 1)(bq+1 - 1)(cr+1 - 1)]/[(a-1)(b-1)(c-1)]
Note:
If a number n can be expressed as n = ap * bq * cr, where a, b, c are primes, then, No. of divisors (factors) of n = (p+1)(q+1)(r+1).
And the sum of the divisors = [(ap+1 - 1)(bq+1 - 1)(cr+1 - 1)]/[(a-1)(b-1)(c-1)]
4.
Answer: C
Explanation:
1) Keeping A in the first position, we can form 4! = 24 words (by arranging the other 4 letters among themselves).
2) Keeping C in first and A in the second position, we can form 3! = 6 words (by arranging the other 3 letters among themselves).
3) Then keeping C in the first, H in the second and A in the third position, we can have 2! = 2 words (by arranging M and S between themselves), namely CHAMS and CHASM, in that order.
Therefore, the position of CHASM in the alphabetically sorted list = 24 + 6 + 2 = 32.
Hence the correct answer is (C).
Explanation:
1) Keeping A in the first position, we can form 4! = 24 words (by arranging the other 4 letters among themselves).
2) Keeping C in first and A in the second position, we can form 3! = 6 words (by arranging the other 3 letters among themselves).
3) Then keeping C in the first, H in the second and A in the third position, we can have 2! = 2 words (by arranging M and S between themselves), namely CHAMS and CHASM, in that order.
Therefore, the position of CHASM in the alphabetically sorted list = 24 + 6 + 2 = 32.
Hence the correct answer is (C).
5.
Answer: D
Explanation:
Between the 2 brothers any of the remaining 18 persons can be placed. Then the rest 17 can be arranged into 17! ways. Again the 2 brothers can interchange their position in 2! = 2 ways. Therefore, the total no. of arrangements in which the 20 persons can be arranged around a circle with one person between the 2 brothers is 2*(18*17!) ways = 2*18! ways.
Explanation:
Between the 2 brothers any of the remaining 18 persons can be placed. Then the rest 17 can be arranged into 17! ways. Again the 2 brothers can interchange their position in 2! = 2 ways. Therefore, the total no. of arrangements in which the 20 persons can be arranged around a circle with one person between the 2 brothers is 2*(18*17!) ways = 2*18! ways.
6.
Answer: b
Explanation:
Explanation:
By Pythagoras, AD = sqrt(40) and BD = sqrt(36 + x2).
Now, we use this in triangle ABD to get 40 + (36 + x2) = (2 + x)2.
On solving, we get x = 18, hence radius = (18 + 2)/2 = 10. Area of semi circle = p(10)2/2 = 50p.
7.
8.
9.
10.
Now, we use this in triangle ABD to get 40 + (36 + x2) = (2 + x)2.
On solving, we get x = 18, hence radius = (18 + 2)/2 = 10. Area of semi circle = p(10)2/2 = 50p.
7.
Correct Choice (C) Answer (150)
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Explanatory Answer
The sum of the 4th and 12thterm = 20.
Let t1 be the first term, t4 be the fourth term, and t12 be the 12thterm
Then t4 + t12 = 20
à t1 + 3d + t1 + 11d = 20
à 2t1 + 14d = 20
à t1 + 7d =120
à t8 = 10.
The sum of the first 15 terms =
In an arithmetic progression t1 + t15 = t2 + t14 = t3 + t13 =... = t8 + t8.
Therefore, the sum of the first 15 terms = 150
-----------------------------------------------------------------
Explanatory Answer
The sum of the 4th and 12thterm = 20.
Let t1 be the first term, t4 be the fourth term, and t12 be the 12thterm
Then t4 + t12 = 20
à t1 + 3d + t1 + 11d = 20
à 2t1 + 14d = 20
à t1 + 7d =120
à t8 = 10.
The sum of the first 15 terms =
In an arithmetic progression t1 + t15 = t2 + t14 = t3 + t13 =... = t8 + t8.
Therefore, the sum of the first 15 terms = 150
8.
Answer:
The cyclicity of 2 is 4 and the unit’s digit of 2 when raised to power of 4n is 6 (e.g., pow(2,4) = 16, pow(2,8) = 256 etc.).
Now 123! is a multiple of 4. Therefore, the unit’s digit of pow(36472,123!) = 6.
Similarly, the cyclicity of 7 ia also 4 and the unit’s digit of 7 when raised to power of 4n is 1.
As 76! is a multiple of 4. Therefore, the unit’s digit of pow(34767,76!) = 1.
Hence the unit’s digit of the given expression is 6*1 = 6.
The cyclicity of 2 is 4 and the unit’s digit of 2 when raised to power of 4n is 6 (e.g., pow(2,4) = 16, pow(2,8) = 256 etc.).
Now 123! is a multiple of 4. Therefore, the unit’s digit of pow(36472,123!) = 6.
Similarly, the cyclicity of 7 ia also 4 and the unit’s digit of 7 when raised to power of 4n is 1.
As 76! is a multiple of 4. Therefore, the unit’s digit of pow(34767,76!) = 1.
Hence the unit’s digit of the given expression is 6*1 = 6.
9.
Explanation:
Let there be x girls and y boys. Then xC2 = 45, hence x = 10. Also yC2 = 190, hence y = 20. No of games with 1 boy and 1 girl = 10 x 20 = 200.
Let there be x girls and y boys. Then xC2 = 45, hence x = 10. Also yC2 = 190, hence y = 20. No of games with 1 boy and 1 girl = 10 x 20 = 200.
10.
Area common to both circles
= 2 x (Area of sector OAB – area of triangle OAB)
= 2 x (1/4 x pr2 – r2) where r = 1.
On solving, we get the answer as p/2 – 1.
