Two APs have the same common difference.
Question: Two APs have the same common difference. The first term of one of these is 1 and that of the other is 8. The difference between their 4th terms is (a) -1 (b) -8 (c) 7 (d) -9 Solution: (c)Let the common difference of two APs are d1and d2, respectively. Bycondition, d1=d2=d (i) Let the first term of first AP (a1) = -1 and the first term of second AP (a2) = 8 We know that, the nth term of an AP, T1= a + (n 1) d 4th term of first AP, T4= a, + (4 1)d = 1 + 3d . and 4th term of second AP, T4...
Read More →The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm.
Question: The dimensions of a metal block are 2.25 m by 1.5 m by 27 cm. It is melted and recast into cubes, each of the side 45 cm. How many cubes are formed? Solution: Dimension of the metal block is $2.25 \mathrm{~m} \times 1.5 \mathrm{~m} \times 27 \mathrm{~cm}$, i. e., $225 \mathrm{~cm} \times 150 \mathrm{~cm} \times 27 \mathrm{~cm}(\because 1 \mathrm{~m}=100 \mathrm{~cm})$. Volume of the metal block $=225 \times 150 \times 27=911250 \mathrm{~cm}^{3}$ This metal block is melted and recast in...
Read More →What is the common difference
Question: What is the common difference of an AP in which $a_{18}-a_{14}=32$ ? (a) 8 (b) $-8$ (c) $-4$ (d) 4 Solution: (a) Given, $a_{18}-a_{14}=32$ $\Rightarrow \quad a+(18-1) d-[a+(14-1) d]=32$ $\left[\because a_{n}=a+(n-1) d\right]$ $\Rightarrow \quad a+17 d-a-13 d=32$ $\Rightarrow \quad 4 d=32$ $\therefore \quad d=8$ which is the required common difference of an AP....
Read More →An 8 m long cuboidal beam of wood when sliced produces four thousand 1cm
Question: An 8 m long cuboidal beam of wood when sliced produces four thousand 1 cm cubes and there is no wastage of wood in this process. If one edge of the beam is 0.5 m, find the third edge. Solution: Length of the wooden beam $=8 \mathrm{~m}$ Width $=0.5 \mathrm{~m}$ Suppose that the height of the beam is $h \mathrm{~m} .$ Then, its volume $=$ length $\times$ width $\times$ height $=8 \times 0.5 \times \mathrm{h}=4 \times \mathrm{hm}^{3}$ Also, it produces 4000 cubes, each of edge $1 \mathrm...
Read More →If the common difference of an
Question: If the common difference of an $\mathrm{AP}$ is 5 , then what is $\mathrm{a}_{18}-\mathrm{a}_{13}$ ? (a) 5 (b) 20 (c) 25 (d) 30 Solution: (c) Given, the common difference of AP i.e., d = 5 Now, $\quad a_{18}-a_{13}=a+(18-1) d-[a+(13-1) d] \quad\left[\because a_{n}=a+(n-1) d\right]$ $=a+17 \times 5-a-12 \times 5$ $=85-60=25$...
Read More →The rainfall on a certain day was 6 cm.
Question: The rainfall on a certain day was 6 cm. How many litres of water fell on 3 hectares of field on that day? Solution: The rainfall on a certain day $=6 \mathrm{~cm}$ $=6 \times \frac{1}{100} \mathrm{~m}(\because 1 \mathrm{~m}=100 \mathrm{~cm})$ $=0.06 \mathrm{~m}$ Area of the field $=3$ hectares We know that 1 hectare $=10000 \mathrm{~m}^{2}$ i.e., 3 hectares $=3 \times 10000 \mathrm{~m}^{2}=30000 \mathrm{~m}^{2}$ Thus, volume of rain water that fell in the field $=($ area of the field $...
Read More →A beam 5 m long and 40 cm wide contains 0.6 cubic metre of wood.
Question: A beam 5 m long and 40 cm wide contains 0.6 cubic metre of wood. How thick is the beam? Solution: Length of the beam $=5 \mathrm{~m}$ Breadth $=40 \mathrm{~cm}$ $=40 \times \frac{1}{100} \mathrm{~m} \quad(\because 100 \mathrm{~cm}=1 \mathrm{~m})$ $=0.4 \mathrm{~m}$ Suppose that the height of the beam is $h \mathrm{~m}$. Also, it is given that the beam contains $0.6$ cubic metre of wood. i.e., volume of the beam $=0.6 \mathrm{~m}^{3}$ Now, volume of the cuboidal beam $=$ length $\times$...
Read More →Which term of an AP : 21,
Question: Which term of an AP : 21, 42, 63, 84, is 210? (a) 9th (b) 10th (c)11th (d) 12th Solution: (b) Let nth term of the given AP be 210 Here, first term, $\quad a=21$ and common difference, $d=42-21=21$ and $a_{n}=210$ $\because$ $a_{n}=a+(n-1) d$ $\Rightarrow \quad 210=21+(n-1) 21$ $\Rightarrow \quad 210=21+21 n-21$ $\Rightarrow \quad 210=21 n \Rightarrow n=10$ Hence, the 10th term of an AP is 210....
Read More →The number of solutions of the system of equations
Question: The number of solutions of the system of equations $2 x+y-z=7$ $x-3 y+2 z=1$ $x+4 y-3 z=5$ is (a) 3 (b) 2 (c) 1 (d) 0 Solution: $(d) 0$ The given system of equations can be written in matrix form as follows:] $\left[\begin{array}{ccc}2 1 -1 \\ 1 -3 2 \\ 1 4 -3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}7 \\ 1 \\ 5\end{array}\right]$ $A X=B$ Here, $A=\left[\begin{array}{ccc}2 1 -1 \\ 1 -3 2 \\ 1 4 -3\end{array}\right], X=\left[\begin{array...
Read More →The number of solutions of the system of equations
Question: The number of solutions of the system of equations $2 x+y-z=7$ $x-3 y+2 z=1$ $x+4 y-3 z=5$ is (a) 3 (b) 2 (c) 1 (d) 0 Solution: $(d) 0$ The given system of equations can be written in matrix form as follows:] $\left[\begin{array}{ccc}2 1 -1 \\ 1 -3 2 \\ 1 4 -3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}7 \\ 1 \\ 5\end{array}\right]$ $A X=B$ Here, $A=\left[\begin{array}{ccc}2 1 -1 \\ 1 -3 2 \\ 1 4 -3\end{array}\right], X=\left[\begin{array...
Read More →If the 2nd term of an AP is 13 and
Question: If the 2nd term of an AP is 13 and 5th term is 25, what is its 7th term? (a) 30 (b) 33 (c) 37 (d) 38 Solution: (b) Given, a2= 13 and a5= 25 $\Rightarrow \quad a+(2-1) d=13 \quad\left[\because a_{n}=a+(n-1) d\right]$ and $\quad a+(5-1) d=25$ $\Rightarrow \quad a+d=13 \quad \ldots$ (i) and $a+4 d=25$ ...(ii) On subtracting Eq. (i) from Eq. (ii), we get $3 d=25-13=12 \Rightarrow d=4$ From Eq. (i), $a=13-4=9$ $\therefore \quad a_{7}=a+(7-1) d=9+6 \times 4=33$...
Read More →A swimming pool is 250 m long and 130 m wide.
Question: A swimming pool is 250 m long and 130 m wide. 3250 cubic metres of water is pumped into it. Find the rise in the level of water. Solution: Length of the pool $=250 \mathrm{~m}$ Breadth of the pool $=130 \mathrm{~m}$ Also, it is given that $3250 \mathrm{~m}^{3}$ of water is poured into it. i.e., volume of water in the pool $=3250 \mathrm{~m}^{3}$ Suppose that the height of the water level is $h \mathrm{~m}$. Then, volume of the water $=$ length $\times$ breadth $\times$ height $\Rightar...
Read More →A rectangular field is 70 m long and 60 m broad.
Question: A rectangular field is 70 m long and 60 m broad. A well of dimensions 14 m 8 m 6 m is dug outside the field and the earth dug-out from this well is spread evenly on the field. How much will the earth level rise? Solution: Dimension of the well $=14 \mathrm{~m} \times 8 \mathrm{~m} \times 6 \mathrm{~m}$ The volume of the dug $-$ out earth $=14 \times 8 \times 6=672 \mathrm{~m}^{3}$ Now, we will spread this dug $-$ out earth on a field whose length, breadth and height are $70 \mathrm{~m}...
Read More →A village, having a population of 4000, requires 150 litres water per head per day.
Question: A village, having a population of 4000, requires 150 litres water per head per day. It has a tank which is 20 m long, 15 m broad and 6 m high. For how many days will the water of this tank last? Solution: A village has population of 4000 and every person needs $150 \mathrm{~L}$ of water a day. So, the total requirement of water in a day $=4000 \times 150 \mathrm{~L}=600000 \mathrm{~L}$ Also, it is given that the length of the water tank is $20 \mathrm{~m}$. Breadth $=15 \mathrm{~m}$ He...
Read More →The 21st term of an AP
Question: The 21st term of an AP whose first two terms are 3 and 4, is (a) 17 (b) 137 (c) 143 (d)-143 Solution: (b) Given, first two terms of an AP are a = 3 and a + d = 4. ⇒ 3 + d = 4...
Read More →The first four terms of an AP
Question: The first four terms of an AP whose first term is 2 and the common difference is-2 are (a) -2,0,2, 4 (b) -2, 4, -8,16 (c) -2,-4,-6,-8 (d) -2, 4, -8, -16 Solution: (c) Let the first four terms of an AP are a, a + d, a + 2d and a + 3d. Given, that first term, a = 2 and common difference, d = 2, then we have an AP as follows -2, 2 2, 2 + 2(-2), 2 + 3(-2) = 2, 4, 6, 8...
Read More →How many bricks each of size 25 cm × 10 cm × 8 cm will be required to build a wall 5 m long,
Question: How many bricks each of size 25 cm 10 cm 8 cm will be required to build a wall 5 m long, 3 m high and 16 cm thick, assuming that the volume of sand and cement used in the construction is negligible? Solution: Dimension of a brick $=25 \mathrm{~cm} \times 10 \mathrm{~cm} \times 8 \mathrm{~cm}$ Volume of a brick $=25 \mathrm{~cm} \times 10 \mathrm{~cm} \times 8 \mathrm{~cm}$ $=2000 \mathrm{~cm}^{3}$ Also, it is given that the length of the wall is $5 \mathrm{~m}$ $=5 \times 100 \mathrm{~...
Read More →Three coins are tossed simultaneously.
Question: Three coins are tossed simultaneously. What is the probability of getting exactly two heads? (i) $\frac{1}{2}$ (ii) $\frac{1}{4}$ (iii) $\frac{3}{8}$ (iv) $\frac{3}{4}$ Solution: (C) $\frac{3}{8}$ Explanation:When 3 coins are tossed simultaneously, the possible outcomes areHHH,HHT,HTH,THH,THT, HTT, TTHandTTT.Total number of possible outcomes = 8 Let E be the event of getting exactly two heads. Then, the favourable outcomes are HHT, THH, and HTH.Number of favourable outcomes = 3 $\there...
Read More →Three coins are tossed simultaneously.
Question: Three coins are tossed simultaneously. What is the probability of getting exactly two heads? (i) $\frac{1}{2}$ (ii) $\frac{1}{4}$ (iii) $\frac{3}{8}$ (iv) $\frac{3}{4}$ Solution: (C) $\frac{3}{8}$ Explanation:When 3 coins are tossed simultaneously, the possible outcomes areHHH,HHT,HTH,THH,THT, HTT, TTHandTTT.Total number of possible outcomes = 8 Let E be the event of getting exactly two heads. Then, the favourable outcomes are HHT, THH, and HTH.Number of favourable outcomes = 3 $\there...
Read More →The 11 th term of an AP -5,
Question: The 11 th term of an AP $-5, \frac{-5}{2}, 0, \frac{5}{2} \ldots$ (a)-20 (b) 20 (c) $-30$ (d) 30 Solution: Given AP, $5, \frac{-5}{2}, 0, \frac{5}{2}$ Here, $a=-5, d=\frac{-5}{2}+5=\frac{5}{2}$ $\therefore \quad a_{11}=a+(11-1) d \quad\left[\because a_{n}=a+(n-1) d\right]$ $=-5+(10) \times \frac{5}{2}$ $=-5+25=20$...
Read More →The system of equation
Question: The system of equation $x+y+z=2,3 x-y+2 z=6$ and $3 x+y+z=-18$ has (a) a unique solution(b) no solution(c) an infinite number of solutions(d) zero solution as the only solution Solution: (a) a unique solution The given system of equations can be written in matrix form as follows: $\left[\begin{array}{ccc}1 1 1 \\ 3 -1 2 \\ 3 1 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}2 \\ 6 \\ -18\end{array}\right]$ $A X=B$ Here, $A=\left[\begin{array...
Read More →The system of equation
Question: The system of equation $x+y+z=2,3 x-y+2 z=6$ and $3 x+y+z=-18$ has (a) a unique solution(b) no solution(c) an infinite number of solutions(d) zero solution as the only solution Solution: (a) a unique solution The given system of equations can be written in matrix form as follows: $\left[\begin{array}{ccc}1 1 1 \\ 3 -1 2 \\ 3 1 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}2 \\ 6 \\ -18\end{array}\right]$ $A X=B$ Here, $A=\left[\begin{array...
Read More →The list of numbers – 10,
Question: The list of numbers 10, 6, 2, 2, is (a) an AP with d = -16 (b) an AP with d = 4 (c) Fan AP with d = 4 (d) not an AP Solution: (b)The given numbers are -10,-6,-2, 2.. Here, a., = -10, a2= -6, a3= -2 and a4= 2 Since, $\quad a_{2}-a_{1}=-6-(-10)$ $=-6+10=4$ $a_{3}-a_{2}=-2-(-6)$ $=-2+6=4$ $a_{4}-a_{3}=2-(-2)$ $=2+2=4$ Each successive term of given list has same difference i.e., 4. So, the given list forms an AP with common difference, d=4....
Read More →Two coins are tossed simultaneously. What is the probability of getting at most one head?
Question: Two coins are tossed simultaneously. What is the probability of getting at most one head? (a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (C) $\frac{2}{3}$ (d) $\frac{3}{4}$ Solution: (d) $\frac{3}{4}$ Explanation:When two coins are tossed simultaneously, the possible outcomes are HH, HT, TH and TT.Total number of possible outcomes = 4 Let E be the event of getting at most one head. Then, the favourable outcomes are HT, TH and TT.Number of favourable outcomes = 3 $\therefore P($ getting at most 1...
Read More →How many planks each of which is 3 m long,
Question: How many planks each of which is 3 m long, 15 cm broad and 5 cm thick can be prepared from a wooden block 6 m long, 75 cm broad and 45 cm thick? Solution: Length of the wooden block $=6 \mathrm{~m}$ $=6 \times 100 \mathrm{~cm} \quad(\because 1 \mathrm{~m}=100 \mathrm{~cm})$ $=600 \mathrm{~cm}$ Breadth of the block $=75 \mathrm{~cm}$ Height of the block $=45 \mathrm{~cm}$ Volume of block $=$ length $\times$ breadth $\times$ height $=600 \times 75 \times 45 $=2025000 \mathrm{~cm}^{3}$ Ag...
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