Solution: (e)

Sum of five numbers = 5 \(\displaystyle \times \) 49 = 245

Sum of first and second numbers = 2 \(\displaystyle \times \) 48 = 96

Sum of fourth and fifth numbers = 2 \(\displaystyle \times \) 48 = 56

Third number = 245 \(\displaystyle -\) 152 = 93

or

Third number = 5 \(\displaystyle \times \) 49 \(\displaystyle -\) 2 \(\displaystyle \times \) 48 \(\displaystyle -\) 2 \(\displaystyle \times \) 28

= 245 \(\displaystyle -\) 96 \(\displaystyle -\) 56 = 93

Alternate method

Average of 5 no’s =49

Therefore, sum of 5 no’s = 5 \(\displaystyle \times \)49 =245

Average of first and second Number =48

Therefore, sum of first and second Number = 2 \(\displaystyle \times \) 48 =96

Average of 4th and 5th number = 28

Therefore, sum of 4th and 5th Number =2 \(\displaystyle \times \)28 = 56

Let the third number =x

Sum of 5 numbers = 96+x+56

Therefore,

96+x+56 =245

x+152=245

x = 245 \(\displaystyle -\) 152

x=93

Solution: (c)

Let the even numbers A, B, C and D be x, x + 2, x + 4 and x+ 6 respectively.

Now, according to the question,

x + x + 2 + x + 4 + x + 6 = 4 \(\displaystyle \times \) 37

\(\displaystyle \Rightarrow \)4x + 12 = 148

\(\displaystyle \Rightarrow \)4x = 148 – 12 = 136

\(\displaystyle \Rightarrow \)\(\displaystyle x=\frac{{136}}{4}=34\)

Therefore, A = 34 and C = (34 + 4) = 38

A \(\displaystyle \times \) C = 34 \(\displaystyle \times \) 38 = 1292

Alternate Method :

Average of consecutive even or odd numbers = a + (n–1), where a is the smallest number.

So, average of four consecutive even numbers = a + (4 – 1)

or, 37 = a + 3

or, a = 37 – 3 = 34

Therefore, A = 34, B = 36, C = 38, D = 40

\(\displaystyle \Rightarrow \) Product of A and C = 34 \(\displaystyle \times \) 38 = 1292

Solution: (b)

Let the sum of maximum marks of all the subjects be ‘x’:

or, \(\displaystyle \frac{{x\times 85}}{{100}}=(103+111+98+110+88)\)

or, x = \(\displaystyle \frac{{510\times 100}}{{85}}=600\)

\(\displaystyle \Rightarrow \)Maximum marks of each subject = \(\displaystyle \frac{{600}}{5}=120\)

Solution: (d)

Let the four odd consecutive numbers be

x, x + 2, x + 4, and x + 6.

Also, A = x, B = x + 2, C = x + 4 and D = x + 6

Therefore, 4x + 12 = 4 × 40

or, 4x = 160 – 12 = 148

or, x = \(\displaystyle \frac{{148}}{4}=37\)

Therefore, Numbers are A = 37, B = 39, C = 41, D = 43

\(\displaystyle \Rightarrow \) Product of B and D = 39 \(\displaystyle \times \)43 = 1677

Alternative Method

Let the four consecutive odd numbers be
A,  B,  C, D
x+1, x+3, x+5,  x+7
According to questions,

\(\displaystyle \frac{{x+1+x+3+x+5+x+7}}{4}=40\)
\(\displaystyle \frac{{4x+16}}{4}=40\)

\(\displaystyle 4x+16=160\)

\(\displaystyle 4x=144\)

x=36

B=x+3
=36+3=39
D = x + 7 = 36 + 7 = 43

Therefore, required product =B\(\displaystyle \times \)D=39 \(\displaystyle \times \)43=1677

Solution: (e)

Let four consecutive numbers are

A = (x), B = (x +1), C = (x + 2)  and D = (x + 3)

According to question

Average = \(\displaystyle \frac{{(x)+(x+1)+(x+2)+(x+3)}}{4}\)

Þ 56.5 = \(\displaystyle \frac{{4x+6}}{4}\)

Þ 226 = 4x + 6

Þ 4x = 226 – 6 = 220

Therefore, x = \(\displaystyle \frac{{220}}{4}=55\)

\(\displaystyle \Rightarrow \) Product of A and C = (x) \(\displaystyle \times \) (x + 2)

= (55) \(\displaystyle \times \) (55 + 2) = 55 \(\displaystyle \times \) 57 = 3135

Solution: (d)

Let Woman’s age = 16x

Daughter’s age = 3x

Now \(\displaystyle \frac{{(16x+3x)}}{2}=19\)

\(\displaystyle \Rightarrow \) 19x = 38

\(\displaystyle \Rightarrow \) x = 2

Therefore, Daughter’s age = 3 × 2 = 6 years

Solution: (b)

Let the five no. Be \(\displaystyle {{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}},{{x}_{5}}\)

Average of 5 numbers = 61

\(\displaystyle \frac{{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}}}{5}=61\)

\(\displaystyle {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}=305\)

Now, \(\displaystyle \frac{{{{x}_{1}}+{{x}_{3}}}}{2}=69\)

\(\displaystyle {{x}_{1}}+{{x}_{3}}=138\)

Also, \(\displaystyle \frac{{{{x}_{2}}+{{x}_{4}}}}{2}=69\)

\(\displaystyle {{x}_{2}}+{{x}_{4}}=138\)

Now, \(\displaystyle {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}=305\)

\(\displaystyle 138+138+{{x}_{5}}=305\)

\(\displaystyle {{x}_{5}}=305-276\)

\(\displaystyle {{x}_{5}}=29\)